Sequences and Series (JEE Main)

Question 1

In an arithmetic progression a₁, a₂, a₃, … it is given that a₃ + a₇ + a₁₂ + a₁₇ + a₂₂ + a₂₆ = 270. The sum of the first 28 terms of the progression is

Accepted Answer

C. Key insight: in an AP, aₚ + a_q depends only on p + q. Here 3 + 26 = 7 + 22 = 12 + 17 = 29 = 1 + 28.
So the given sum is three equal pairs: 3(a₁ + a₂₈) = 270, giving a₁ + a₂₈ = 90.
Hence S₂₈ = 28/2(a₁ + a₂₈) = 14 × 90 = 1260.
Traps: 1350 comes from taking 15 pairs (computing S₃₀); 2520 comes from forgetting the factor 1/2 in Sₙ.
Exam pattern: JEE Main – sum of AP from sum of selected terms with eq

Question 2

For an arithmetic progression, the sum of the first 7 terms is 11 and the sum of the first 11 terms is 7. Find the value of a₈ + a₉ + a₁₀ + … + a₁₈.

Accepted Answer

-29. Key insight: if Sₘ = n and Sₙ = m (m ≠ n), then Sₘ₊ₙ = −(m+n).
Proof: Sₖ = Ak² + Bk. Subtracting Am² + Bm = n and An² + Bn = m gives A(m+n) + B = −1, so Sₘ₊ₙ = (m+n)[A(m+n) + B] = −(m+n).
Thus S₁₈ = −18.
Required sum = S₁₈ − S₇ = −18 − 11 = −29.
Trap: answering S₁₈ = −18 itself, or subtracting S₁₁ instead of S₇.
Exam pattern: JEE Main – AP with S_p = q and S_q = p

Question 3

The sums of the first n terms of two arithmetic progressions are in the ratio (2n + 3) : (3n + 1) for every natural number n. The ratio of the 6th term of the first progression to the 9th term of the

Accepted Answer

B. Key insight: since Sₙ of an AP has the form pn² + qn, the ratio condition forces Sₙ = kn(2n + 3) and Tₙ = kn(3n + 1) with the same constant k.
Then aₘ = Sₘ − S_(m−1) = k[2(2m − 1) + 3] and b_j = k[3(2j − 1) + 1].
a₆ = k(2 · 11 + 3) = 25k, b₉ = k(3 · 17 + 1) = 52k, so the ratio is 25 : 52.
Trap: the shortcut 'replace n by 2m − 1' works only for equal indices; 25 : 34 is the ratio of the 6th terms a

Sequences and Series: questions with solutions

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