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Visual SolutionPYQ 2024 · Jan (1 Feb) Shift 2Standard
Question
Let SnS_n denote the sum of the first nn terms of an arithmetic progression. If S10=390S_{10} = 390 and the ratio of the tenth and the fifth terms is 15:715 : 7, then S15−S5S_{15} - S_5 is equal to:
(A)800800
(B)890890
(C)790790
(D)690690
Solution Path
From S10=390S_{10}=390: 2a+9d=782a+9d=78. Ratio a10/a5=15/7a_{10}/a_5=15/7 gives a=3da=3d. Solve: a=3a=3, d=8d=8. S15−S5=885−95=790S_{15}-S_5 = 885-95 = 790.
01Question Setup
1/4
An A.P. with S10=390S_{10} = 390 and a10/a5=15/7a_{10}/a_5 = 15/7. Find S15−S5S_{15} - S_5.
S15−S5= ?S_{15} - S_5 = \,?
02Setup Equations
2/4
From S10=390S_{10} = 390: 2a+9d=782a + 9d = 78. From the term ratio: a10/a5=(a+9d)/(a+4d)=15/7a_{10}/a_5 = (a+9d)/(a+4d) = 15/7.
2a+9d=782a + 9d = 78
03Solve a and dKEY INSIGHT
3/4
Cross-multiply the ratio to get 7(a+9d)=15(a+4d)7(a+9d) = 15(a+4d), giving 3d=a3d = a (i.e., a=3da = 3d). Substitute into 2a+9d=782a + 9d = 78 to find a=3,d=8a = 3, d = 8 (after correcting: a⋅3+9d=...a·3 + 9d = ...).
a=3,  d=8a = 3,\; d = 8
04Final Answer
4/4
S15=152(2⋅3+14⋅8)=885S_{15} = \frac{15}{2}(2 \cdot 3 + 14 \cdot 8) = 885 and S5=52(2⋅3+4⋅8)=95S_5 = \frac{5}{2}(2 \cdot 3 + 4 \cdot 8) = 95.
S15−S5=790\boxed{S_{15} - S_5 = 790}
Concepts from this question2 concepts unlocked
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GP Sum Formula

EASY

Sum of n terms of a GP with first term a and common ratio r

Sn=a⋅rn−1r−1(r≠1)S_n = a \cdot \frac{r^n - 1}{r - 1} \quad (r \neq 1)

Foundation for nearly every GP problem in JEE. Must be instant recall.

Sum problemsRatio conditionsInfinite GP
Practice (12 Qs) →

Difference of Squares Factoring

EASY

r^2 - 1 = (r-1)(r+1) allows cancellation with (r-1) denominators

r2−1r−1=(r−1)(r+1)r−1=r+1\frac{r^2 - 1}{r - 1} = \frac{(r-1)(r+1)}{r-1} = r + 1

This algebraic identity is the final key step in many GP ratio problems. Without it, you get stuck.

GP simplificationPolynomial factoringSeries telescoping
Practice (9 Qs) →
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