Arithmetic Progressions Assertion–Reason PYQs Class 10 CBSE (With Answers)

Official CBSE answer · Explanation by Team TutorVertex

Assertion: True.
If a, b, c are in AP, b − a = c − b.
Therefore, 2b = a + c.
Conversely, 2b = a + c gives b − a = c − b, so the terms are in AP.

Reason: True.
For 1, 3, 5, …, (2n − 1), the first term is 1 and common difference is 2.
Sₙ = $\frac{\text{n}}{\text{2}}$[2a + (n − 1)d]
= $\frac{\text{n}}{\text{2}}$[2 + 2(n − 1)]
= n².

Link: R gives a sum of odd numbers. It does not explain the condition for the three given terms to be in AP. Both statements are true, but R does not explain A. Answer: B.

Official CBSE answer · Explanation by Team TutorVertex

Assertion: False.
d = second term − first term
= 1 − 5
= −4.
Also, −3 − 1 = −4 and −7 − (−3) = −4.

Reason: True.
For successive terms of an AP, the common difference is later term − preceding term.
Thus d = aₙ − aₙ₋₁ for n ≥ 2.

Link: The rule in R gives −4, which contradicts the value 4 in A. A is false and R is true. Answer: D.

Official CBSE answer · Explanation by Team TutorVertex

Assertion: True. The displayed fractions require m ≠ 0.
a = $\frac{\text{1}}{\text{m}}$
d = $\frac{\text{1 + 2m}}{\text{m}}$ − $\frac{\text{1}}{\text{m}}$ = 2
a₈ = a + (8 − 1)d
= $\frac{\text{1}}{\text{m}}$ + 14
= $\frac{\text{1 + 14m}}{\text{m}}$.

Reason: True.
The nth term is obtained by adding (n − 1) common differences to the first term.
Hence aₙ = a + (n − 1)d.

Link: Substituting a = $\frac{\text{1}}{\text{m}}$, d = 2 and n = 8 into R gives the eighth term stated in A. R explains A. Answer: A.

Official CBSE answer · Explanation by Team TutorVertex

Assertion: True.
a = 21, d = 18 − 21 = −3 and n = 15.
Sₙ = $\frac{\text{n}}{\text{2}}$[2a + (n − 1)d]
S₁₅ = $\frac{\text{15}}{\text{2}}$[2(21) + 14(−3)]
= $\frac{\text{15}}{\text{2}}$(42 − 42)
= 0.

Reason: False.
The correct formula is Sₙ = $\frac{\text{n}}{\text{2}}$[2a + (n − 1)d].
The expression printed in R has a in place of 2a.

Link: A gives the correct sum, but R states an incorrect general formula. A is true and R is false. Answer: C.

Official CBSE answer · Explanation by Team TutorVertex

Assertion: False.
l = 198, d = 3 and n = 10.
The nth term from the end = l − (n − 1)d.
The 10th term from the end = 198 − (10 − 1)(3)
= 198 − 27
= 171, which is not 168.

Reason: True.
Reading the AP backwards gives first term l and common difference −d.
Its nth term is l + (n − 1)(−d) = l − (n − 1)d.

Link: R gives 171 as the 10th term from the end, so it disproves the value stated in A. A is false and R is true. Answer: D.

Official CBSE answer · Explanation by Team TutorVertex

Assertion: True.
a₂ − a₁ = $\frac{\text{−5}}{\text{2}}$ − (−5) = $\frac{\text{5}}{\text{2}}$
a₃ − a₂ = 0 − ($\frac{\text{−5}}{\text{2}}$) = $\frac{\text{5}}{\text{2}}$
a₄ − a₃ = $\frac{\text{5}}{\text{2}}$ − 0 = $\frac{\text{5}}{\text{2}}$.
The displayed pattern has common difference $\frac{\text{5}}{\text{2}}$, so it is an AP.

Reason: False.
The same AP contains negative rational numbers −5 and $\frac{\text{−5}}{\text{2}}$, as well as the positive rational number $\frac{\text{5}}{\text{2}}$.

Link: The common difference establishes A, while this sequence is a counterexample to R. A is true and R is false. Answer: C.