The common difference of the A.P. a1, a2, …, am is 13 more than the common difference of the A.P. b1, b2, …, bn. If b31 = −277, b43 = −385 and a78 = 327, then a1 is equal to
Q. The common difference of the A.P.:
a1, a2, … , am is 13 more than the common difference of the A.P.:
b1, b2, … , bn. If

b31 = −277,   b43 = −385 and a78 = 327,
then a1 is equal to

(A) 21

(B) 19

(C) 24

(D) 16

Correct Answer: 19

Explanation

Let the common difference of A.P. b be d. Then the common difference of A.P. a is d + 13.

Using the formula of nth term:

bn = b1 + (n − 1)d

From given data:

b31 = b1 + 30d = −277

b43 = b1 + 42d = −385

Subtracting the two equations:

12d = −108 ⇒ d = −9

So, common difference of A.P. a is:

d + 13 = −9 + 13 = 4

Now use:

a78 = a1 + 77 × 4

327 = a1 + 308

a1 = 19

Hence, the value of a1 is 19.

Related JEE Main Questions

For three unit vectors a, b, c satisfying |a − b|² + |b − c|² + |c − a|² = 9 and |2a + kb + kc| = 3, the positive value of k is
If ∫ (1 − 5cos²x)/(sin⁵x cos²x) dx = f(x) + C, where C is the constant of integration, then f(π/6) − f(π/4) is equal to
Let S = {x³ + ax² + bx + c : a, b, c ∈ N and a, b, c ≤ 20} be a set of polynomials. Then the number of polynomials in S, which are divisible by x² + 2, is
X is the number of geometrical isomers exhibited by [Pt(NH3)(H2O)BrCl]. Y is the number of optically inactive isomer(s) exhibited by [CrCl2(ox)2]3−. Z is the number of geometrical isomers exhibited by [Co(NH3)3(NO2)3]. The value of X + Y + Z is ____ .
Consider the dissociation equilibrium of the following weak acid HA ⇌ H⁺(aq) + A⁻(aq). If the pKa of the acid is 4, then the pH of 10 mM HA solution is ____ . (Nearest integer)
JEE Main Chapter-wise Previous Year Questions

Updated for JEE Main 2026: This PYQ is important for JEE Mains, JEE Advanced and other competitive exams. Practice more questions from this chapter.

Scroll to Top