Arithmetic Progressions 2-Mark Questions Class 10: CBSE PYQs with Answers

a = 3 and d = 8 − 3 = 5.

aₙ = a + (n − 1)d

78 = 3 + 5(n − 1)

75 = 5(n − 1)

n = 16.

78 is the 16th term

a₁ = 6(1) − 5 = 1.

a₂ = 6(2) − 5 = 7.

d = a₂ − a₁ = 7 − 1 = 6.

d = 6

a = $\frac{\text{(}\text{11}\text{)/(}}{\text{4}\text{)}}$ and d = $\frac{\text{(}\text{7}\text{)/(}}{\text{2}\text{)}}$ − $\frac{\text{(}\text{11}\text{)/(}}{\text{4}\text{)}}$ = $\frac{\text{(}\text{3}\text{)/(}}{\text{4}\text{)}}$.

aₙ = a + (n − 1)d

$\frac{\text{(}\text{53}\text{)/(}}{\text{4}\text{)}}$ = $\frac{\text{(}\text{11}\text{)/(}}{\text{4}\text{)}}$ + (n − 1)$\frac{\text{(}\text{3}\text{)/(}}{\text{4}\text{)}}$

53 = 11 + 3(n − 1)

n − 1 = 14

n = 15.

$\frac{\text{(}\text{53}\text{)/(}}{\text{4}\text{)}}$ is the 15th term

a = −3 and d = 4 − (−3) = 7.

aₙ = a + (n − 1)d

a₃₆ = −3 + 35 × 7

a₃₆ = −3 + 245 = 242.

a₃₆ = 242

√27 = 3√3, √48 = 4√3 and √75 = 5√3.

d = 4√3 − 3√3 = √3.

Next term = 5√3 + √3 = 6√3.

Following term = 6√3 + √3 = 7√3.

Next two terms: 6√3 and 7√3

The original common difference is d = 8 − 3 = 5.

The kth term from the end is l − (k − 1)d.

15th term from the end = 253 − 14 × 5

= 253 − 70 = 183.

15th term from the end = 183

√5, √20 and √45 are √5, 2√5 and 3√5.

d = 2√5 − √5 = √5.

Next term = 3√5 + √5 = 4√5.

Following term = 4√5 + √5 = 5√5.

Next two terms: 4√5 and 5√5

a = 2 and d = 11 − 2 = 9.

aₙ − a₂₅ = (n − 25)d

99 = 9(n − 25)

n − 25 = 11

n = 36.

Required term: 36th

The odd numbers are 21, 23, …, 69.

a = 21, d = 2 and aₙ = 69.

aₙ = a + (n − 1)d

69 = 21 + 2(n − 1)

n = 25.

25 odd numbers

a = −$\frac{\text{(}\text{11}\text{)/(}}{\text{2}\text{)}}$ and d = −3 − (−$\frac{\text{(}\text{11}\text{)/(}}{\text{2}\text{)}}$) = $\frac{\text{(}\text{5}\text{)/(}}{\text{2}\text{)}}$.

aₙ = a + (n − 1)d

$\frac{\text{(}\text{49}\text{)/(}}{\text{2}\text{)}}$ = −$\frac{\text{(}\text{11}\text{)/(}}{\text{2}\text{)}}$ + (n − 1)$\frac{\text{(}\text{5}\text{)/(}}{\text{2}\text{)}}$

49 = −11 + 5(n − 1)

n − 1 = 12

n = 13.

$\frac{\text{(}\text{49}\text{)/(}}{\text{2}\text{)}}$ is the 13th term

Let the common difference be d.

23 − 7 = 2d

d = 8.

a = 7 − d = 7 − 8 = −1.

b = 7 + d = 7 + 8 = 15.

a = −1, b = 15

For the first AP, aₙ = 9 + (n − 1)(−2).

For the second AP, bₙ = 15 + (n − 1)(−3).

9 − 2(n − 1) = 15 − 3(n − 1)

11 − 2n = 18 − 3n

n = 7.

Check: a₇ = 9 − 12 = −3 and b₇ = 15 − 18 = −3.

n = 7; both terms are −3

Numbers divisible by 5 but not by 2 are 5, 15, 25, …, 995.

a = 5, d = 10 and aₙ = 995.

aₙ = a + (n − 1)d

995 = 5 + 10(n − 1)

n = 100.

100 natural numbers

a = 5, d = 11 − 5 = 6 and aₙ = 203.

aₙ = a + (n − 1)d

203 = 5 + 6(n − 1)

n − 1 = 33

n = 34.

34 terms

a₁ = 5 − 3(1) = 2.

a₂₀ = 5 − 3(20) = −55.

Sₙ = $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$(a + l)

S₂₀ = $\frac{\text{(}\text{20}\text{)/(}}{\text{2}\text{)}}$(2 − 55)

S₂₀ = 10 × (−53) = −530.

S₂₀ = −530

a₁ = $\frac{\text{(}\text{1 × 2 × 3}\text{)/(}}{\text{6}\text{)}}$ = 1.

a₂ = $\frac{\text{(}\text{2 × 3 × 5}\text{)/(}}{\text{6}\text{)}}$ = 5.

a₃ = $\frac{\text{(}\text{3 × 4 × 7}\text{)/(}}{\text{6}\text{)}}$ = 14.

a₂ − a₁ = 5 − 1 = 4.

a₃ − a₂ = 14 − 5 = 9.

The successive differences are unequal, so the sequence is not an AP.

First three terms: 1, 5, 14; not an AP

aₙ = 2n + 3.

a₆ = 2(6) + 3 = 15.

a₂₀ = 2(20) + 3 = 43.

a₆ = 15, a₂₀ = 43

a = 4, d = 10 − 4 = 6 and aₙ = 208.

aₙ = a + (n − 1)d

208 = 4 + 6(n − 1)

n − 1 = 34

n = 35.

35 terms

The multiples form the AP 7, 14, 21, … .

a = 7, d = 7 and n = 15.

Sₙ = $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$[2a + (n − 1)d]

S₁₅ = $\frac{\text{(}\text{15}\text{)/(}}{\text{2}\text{)}}$[14 + 14 × 7]

S₁₅ = $\frac{\text{(}\text{15}\text{)/(}}{\text{2}\text{)}}$ × 112 = 840.

Sum = 840

a = 293, d = 285 − 293 = −8 and aₙ = 53.

aₙ = a + (n − 1)d

53 = 293 − 8(n − 1)

8(n − 1) = 240

n = 31.

31 terms

The required integers form the AP 7, 14, 21, … .

a = 7, d = 7 and n = 40.

Sₙ = $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$[2a + (n − 1)d]

S₄₀ = $\frac{\text{(}\text{40}\text{)/(}}{\text{2}\text{)}}$[14 + 39 × 7]

S₄₀ = 20 × 287 = 5740.

Sum = 5740

Here a = 12x, d = −2x and l = −2x.

l = a + (n − 1)d

−2x = 12x − 2x(n − 1)

For x ≠ 0, dividing by −2x gives 1 = −6 + n − 1.

n = 8.

This is the non-zero-x reading used in the scheme. If x = 0, every displayed term is zero and the endpoint does not determine the number of terms.

8 terms, for x ≠ 0

The integers are 3, 6, 9, …, 99.

a = 3, d = 3 and l = 99.

l = a + (n − 1)d

99 = 3 + 3(n − 1)

n = 33.

33 integers

d = 15 − 5 = 10.

aₙ − a₃₁ = (n − 31)d

130 = 10(n − 31)

n − 31 = 13

n = 44.

Required term: 44th

d = 12 − 6 = 6.

aₙ − a₂₅ = (n − 25)d

150 = 6(n − 25)

n − 25 = 25

n = 50.

Required term: 50th

a₅ = a + 4d = 3.

a₁₂ = a + 11d = 17.

Subtracting gives 7d = 14, so d = 2.

a = 3 − 4 × 2 = −5.

aₙ = a + (n − 1)d

aₙ = −5 + 2(n − 1) = 2n − 7.

aₙ = 2n − 7

Let the first term be a. Here d = 6 and S₁₀ = 4S₅.

Sₙ = $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$[2a + (n − 1)d]

$\frac{\text{(}\text{10}\text{)/(}}{\text{2}\text{)}}$(2a + 9 × 6) = 4 × $\frac{\text{(}\text{5}\text{)/(}}{\text{2}\text{)}}$(2a + 4 × 6)

10a + 270 = 20a + 240

10a = 30

a = 3.

First term = 3

a = 17 and a₄ = 44.

a₄ = a + 3d

44 = 17 + 3d

d = 9.

a₁₅ = a + 14d

a₁₅ = 17 + 14 × 9 = 143.

a₁₅ = 143

a = −24 and a₁₁ = 21.

a₁₁ = a + 10d

21 = −24 + 10d

d = $\frac{\text{(}\text{45}\text{)/(}}{\text{10}\text{)}}$ = $\frac{\text{(}\text{9}\text{)/(}}{\text{2}\text{)}}$.

a₈ = a + 7d

a₈ = −24 + 7 × $\frac{\text{(}\text{9}\text{)/(}}{\text{2}\text{)}}$ = $\frac{\text{(}\text{15}\text{)/(}}{\text{2}\text{)}}$ = 7.5.

a₈ = $\frac{\text{(}\text{15}\text{)/(}}{\text{2}\text{)}}$ = 7.5

a₃ + a₇ = 0.

(a + 2d) + (a + 6d) = 0

2(a + 4d) = 0

a₅ = a + 4d = 0.

a₅ = 0

a + 2d = 5.

a + 6d = 9.

Subtracting gives 4d = 4.

d = 1.

a = 5 − 2 × 1 = 3.

Hence the AP is 3, 4, 5, 6, … .

AP: 3, 4, 5, 6, …

a = 10, n = 14 and S₁₄ = 1505.

Sₙ = $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$[2a + (n − 1)d]

1505 = $\frac{\text{(}\text{14}\text{)/(}}{\text{2}\text{)}}$(20 + 13d)

215 = 20 + 13d

13d = 195

d = 15.

d = 15

For three consecutive terms of an AP, the successive differences are equal.

k − $\frac{\text{(}\text{4}\text{)/(}}{\text{5}\text{)}}$ = 2 − k

2k = 2 + $\frac{\text{(}\text{4}\text{)/(}}{\text{5}\text{)}}$ = $\frac{\text{(}\text{14}\text{)/(}}{\text{5}\text{)}}$

k = $\frac{\text{(}\text{7}\text{)/(}}{\text{5}\text{)}}$.

k = $\frac{\text{(}\text{7}\text{)/(}}{\text{5}\text{)}}$

$\frac{\text{(}\text{a₄}\text{)/(}}{\text{a₇}\text{)}}$ = $\frac{\text{(}\text{a + 3d}\text{)/(}}{\text{a + 6d}\text{)}}$ = $\frac{\text{(}\text{2}\text{)/(}}{\text{3}\text{)}}$.

3(a + 3d) = 2(a + 6d)

a = 3d.

$\frac{\text{(}\text{a₆}\text{)/(}}{\text{a₈}\text{)}}$ = $\frac{\text{(}\text{a + 5d}\text{)/(}}{\text{a + 7d}\text{)}}$

= $\frac{\text{(}\text{3d + 5d}\text{)/(}}{\text{3d + 7d}\text{)}}$ = $\frac{\text{(}\text{8d}\text{)/(}}{\text{10d}\text{)}}$ = $\frac{\text{(}\text{4}\text{)/(}}{\text{5}\text{)}}$.

The given ratio requires a₇ ≠ 0; with a = 3d this also gives d ≠ 0.

$\frac{\text{(}\text{a₆}\text{)/(}}{\text{a₈}\text{)}}$ = $\frac{\text{(}\text{4}\text{)/(}}{\text{5}\text{)}}$

a = 50, d = −4 and Sₙ = 0.

Sₙ = $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$[2a + (n − 1)d]

0 = $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$[100 − 4(n − 1)]

Since n > 0, 104 − 4n = 0.

n = 26.

n = 26

The first twelve 2-digit multiples are 14, 21, …, 91.

a = 14, d = 7 and n = 12.

Sₙ = $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$[2a + (n − 1)d]

S₁₂ = $\frac{\text{(}\text{12}\text{)/(}}{\text{2}\text{)}}$[28 + 11 × 7]

S₁₂ = 6 × 105 = 630.

Sum = 630

The multiples form the AP 8, 16, 24, … .

a = 8, d = 8 and n = 15.

Sₙ = $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$[2a + (n − 1)d]

S₁₅ = $\frac{\text{(}\text{15}\text{)/(}}{\text{2}\text{)}}$[16 + 14 × 8]

S₁₅ = $\frac{\text{(}\text{15}\text{)/(}}{\text{2}\text{)}}$ × 128 = 960.

Sum = 960

a = −30, d = −24 − (−30) = 6 and n = 30.

Sₙ = $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$[2a + (n − 1)d]

S₃₀ = $\frac{\text{(}\text{30}\text{)/(}}{\text{2}\text{)}}$[−60 + 29 × 6]

S₃₀ = 15 × 114 = 1710.

S₃₀ = 1710

Sₙ = n(4n + 1) = 4n² + n.

a₁ = S₁ = 4 + 1 = 5.

S₂ = 4(2)² + 2 = 18.

a₂ = S₂ − S₁ = 18 − 5 = 13.

d = a₂ − a₁ = 13 − 5 = 8.

Hence the AP is 5, 13, 21, 29, … .

AP: 5, 13, 21, 29, …

The first twelve 2-digit multiples are 12, 18, …, 78.

a = 12, d = 6 and n = 12.

Sₙ = $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$[2a + (n − 1)d]

S₁₂ = $\frac{\text{(}\text{12}\text{)/(}}{\text{2}\text{)}}$[24 + 11 × 6]

S₁₂ = 6 × 90 = 540.

Sum = 540

a₂ = a + d = 26.

a₁₅ = a + 14d = −26.

Subtracting gives 13d = −52.

d = −4.

a = 26 − (−4) = 30.

Hence the AP is 30, 26, 22, 18, … .

AP: 30, 26, 22, 18, …

a₁ = 5 − 2(1) = 3.

a₂₀ = 5 − 2(20) = −35.

Sₙ = $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$(a + l)

S₂₀ = $\frac{\text{(}\text{20}\text{)/(}}{\text{2}\text{)}}$[3 + (−35)]

S₂₀ = 10 × (−32) = −320.

S₂₀ = −320

d = 5 and a₂₀ = 135.

a₂₀ = a + 19d

135 = a + 19 × 5

a = 40.

S₂₀ = $\frac{\text{(}\text{20}\text{)/(}}{\text{2}\text{)}}$(a + a₂₀)

S₂₀ = 10(40 + 135) = 1750.

S₂₀ = 1750

a = 5, l = 15 and Sₙ = 30.

Sₙ = $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$(a + l)

30 = $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$(5 + 15)

30 = 10n

n = 3.

n = 3

a₆ = a + 5d = 30.

S₁₁ = $\frac{\text{(}\text{11}\text{)/(}}{\text{2}\text{)}}$(2a + 10d)

S₁₁ = 11(a + 5d)

S₁₁ = 11 × 30 = 330.

S₁₁ = 330

a = 5, l = 40 and Sₙ = 180.

Sₙ = $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$(a + l)

180 = $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$(5 + 40)

360 = 45n

n = 8.

n = 8

a = 12 and a₁₁ − a₇ = 24.

(a + 10d) − (a + 6d) = 24

4d = 24, so d = 6.

a₂₀ = a + 19d

a₂₀ = 12 + 19 × 6 = 126.

a₂₀ = 126

The AP is 5, 10, 15, … .

a = 5, d = 5 and n = 10.

Sₙ = $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$[2a + (n − 1)d]

S₁₀ = $\frac{\text{(}\text{10}\text{)/(}}{\text{2}\text{)}}$[10 + 9 × 5]

S₁₀ = 5 × 55 = 275.

Sum = 275

√3, √12 and √27 are √3, 2√3 and 3√3.

a = √3 and d = √3.

aₙ = a + (n − 1)d

a₁₀ = √3 + 9√3 = 10√3.

a₁₀ = 10√3

a = $\frac{\text{(}\text{1}\text{)/(}}{\text{15}\text{)}}$ and d = $\frac{\text{(}\text{1}\text{)/(}}{\text{12}\text{)}}$ − $\frac{\text{(}\text{1}\text{)/(}}{\text{15}\text{)}}$ = $\frac{\text{(}\text{1}\text{)/(}}{\text{60}\text{)}}$.

Sₙ = $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$[2a + (n − 1)d]

S₁₅ = $\frac{\text{(}\text{15}\text{)/(}}{\text{2}\text{)}}$($\frac{\text{(}\text{2}\text{)/(}}{\text{15}\text{)}}$ + 14 × $\frac{\text{(}\text{1}\text{)/(}}{\text{60}\text{)}}$)

S₁₅ = $\frac{\text{(}\text{15}\text{)/(}}{\text{2}\text{)}}$ × $\frac{\text{(}\text{22}\text{)/(}}{\text{60}\text{)}}$ = $\frac{\text{(}\text{11}\text{)/(}}{\text{4}\text{)}}$.

S₁₅ = $\frac{\text{(}\text{11}\text{)/(}}{\text{4}\text{)}}$

a = $\frac{\text{(}\text{2}\text{)/(}}{\text{3}\text{)}}$ and d = 0 − $\frac{\text{(}\text{2}\text{)/(}}{\text{3}\text{)}}$ = −$\frac{\text{(}\text{2}\text{)/(}}{\text{3}\text{)}}$.

Sₙ = $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$[2a + (n − 1)d]

S₂₀ = $\frac{\text{(}\text{20}\text{)/(}}{\text{2}\text{)}}$($\frac{\text{(}\text{4}\text{)/(}}{\text{3}\text{)}}$ + 19(−$\frac{\text{(}\text{2}\text{)/(}}{\text{3}\text{)}}$))

S₂₀ = 10 × (−$\frac{\text{(}\text{34}\text{)/(}}{\text{3}\text{)}}$) = −$\frac{\text{(}\text{340}\text{)/(}}{\text{3}\text{)}}$.

S₂₀ = −$\frac{\text{(}\text{340}\text{)/(}}{\text{3}\text{)}}$

a = −$\frac{\text{(}\text{29}\text{)/(}}{\text{3}\text{)}}$ and d = −9 − (−$\frac{\text{(}\text{29}\text{)/(}}{\text{3}\text{)}}$) = $\frac{\text{(}\text{2}\text{)/(}}{\text{3}\text{)}}$.

Sₙ = $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$[2a + (n − 1)d]

S₂₀ = $\frac{\text{(}\text{20}\text{)/(}}{\text{2}\text{)}}$[−$\frac{\text{(}\text{58}\text{)/(}}{\text{3}\text{)}}$ + 19 × $\frac{\text{(}\text{2}\text{)/(}}{\text{3}\text{)}}$]

S₂₀ = 10 × (−$\frac{\text{(}\text{20}\text{)/(}}{\text{3}\text{)}}$) = −$\frac{\text{(}\text{200}\text{)/(}}{\text{3}\text{)}}$.

S₂₀ = −$\frac{\text{(}\text{200}\text{)/(}}{\text{3}\text{)}}$

a = 32, l = −10 and d = −2.

l = a + (n − 1)d

−10 = 32 − 2(n − 1)

n − 1 = 21, so n = 22.

S₂₂ = $\frac{\text{(}\text{22}\text{)/(}}{\text{2}\text{)}}$[32 + (−10)]

S₂₂ = 11 × 22 = 242.

n = 22, S₂₂ = 242

a₁ = 3(1) − 2 = 1.

a₂₈ = 3(28) − 2 = 82.

Sₙ = $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$(a + l)

S₂₈ = $\frac{\text{(}\text{28}\text{)/(}}{\text{2}\text{)}}$(1 + 82)

S₂₈ = 14 × 83 = 1162.

S₂₈ = 1162

a = 4, l = 31 and Sₙ = 175.

Sₙ = $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$(a + l)

175 = $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$(4 + 31)

n = 10.

l = a + (n − 1)d gives 31 = 4 + 9d.

d = 3.

n = 10, d = 3

a = 21, d = 18 − 21 = −3 and Sₙ = 0.

Sₙ = $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$[2a + (n − 1)d]

0 = $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$[42 − 3(n − 1)]

Since n > 0, 45 − 3n = 0.

n = 15.

15 terms

The first ten 2-digit multiples are 14, 21, …, 77.

a = 14, l = 77 and n = 10.

Sₙ = $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$(a + l)

S₁₀ = $\frac{\text{(}\text{10}\text{)/(}}{\text{2}\text{)}}$(14 + 77)

S₁₀ = 5 × 91 = 455.

Sum = 455

a₃ = a + 2d = 13.

a₈ = a + 7d = 38.

Subtracting gives 5d = 25, so d = 5.

a = 13 − 2 × 5 = 3.

a₁₁ = a + 10d

a₁₁ = 3 + 10 × 5 = 53.

a₁₁ = 53

a = 20, n = 12 and S₁₂ = 900.

Sₙ = $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$[2a + (n − 1)d]

900 = $\frac{\text{(}\text{12}\text{)/(}}{\text{2}\text{)}}$(40 + 11d)

150 = 40 + 11d, so d = 10.

a₁₂ = a + 11d

a₁₂ = 20 + 11 × 10 = 130.

d = 10, a₁₂ = 130

Sₙ = 6n − n².

a₁ = S₁ = 6 − 1 = 5.

S₂ = 6(2) − 2² = 8.

a₂ = S₂ − S₁ = 8 − 5 = 3.

d = a₂ − a₁ = 3 − 5 = −2.

d = −2

a = 8, d = 2 and n = 60.

The last 10 terms run from a₅₁ to a₆₀.

a₅₁ = 8 + 50 × 2 = 108.

a₆₀ = 8 + 59 × 2 = 126.

Sum = $\frac{\text{(}\text{10}\text{)/(}}{\text{2}\text{)}}$(108 + 126)

= 5 × 234 = 1170.

Sum of last 10 terms = 1170

a = 6, d = 13 − 6 = 7 and l = 230.

l = a + (n − 1)d

230 = 6 + 7(n − 1)

n = 33.

The middle term is the $\frac{\text{(}\text{33 + 1}\text{)/(}}{\text{2}\text{)}}$th, that is, the 17th term.

a₁₇ = 6 + 16 × 7 = 118.

Middle term: a₁₇ = 118

aₙ = Sₙ − Sₙ₋₁, for n ≥ 2.

S₁₀ = 2(10)² + 10 = 210.

S₉ = 2(9)² + 9 = 171.

a₁₀ = S₁₀ − S₉

a₁₀ = 210 − 171 = 39.

a₁₀ = 39

aₙ = Sₙ − Sₙ₋₁, for n ≥ 2.

S₂₅ = 4(25)² − 25 = 2475.

S₂₄ = 4(24)² − 24 = 2280.

a₂₅ = S₂₅ − S₂₄

a₂₅ = 2475 − 2280 = 195.

a₂₅ = 195