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Visual SolutionPYQ 2024 · Jan 27 Shift 2Standard
Question
The 20th20^{\text{th}} term from the end of the progression 20,1914,1812,1734,…,−1291420, 19\frac{1}{4}, 18\frac{1}{2}, 17\frac{3}{4}, \ldots, -129\frac{1}{4} is:
(A)−118-118
(B)−110-110
(C)−115-115
(D)−100-100
Solution Path
AP with a=20a=20, d=−3/4d=-3/4, n=200n=200. 20th from end = 181st term =20−135=−115= 20 - 135 = -115.
01Question Setup
1/4
Find the 20th term from the end of an arithmetic progression.
a20 from end=  ?a_{20 \text{ from end}} = \;?
02AP Parameters
2/4
a=20a = 20, d=−34d = -\frac{3}{4}, last term l=−5174l = -\frac{517}{4}. Solve for n=200n = 200 terms.
n=200n = 200
03Term from EndKEY INSIGHT
3/4
20th from end = term 181 from start. a181=20+180(−34)=20−135=−115a_{181} = 20 + 180\left(-\frac{3}{4}\right) = 20 - 135 = -115.
a181=20−135=−115a_{181} = 20 - 135 = -115
04Final Answer
4/4
The 20th term from the end is −115-115.
−115\boxed{-115}
Concepts from this question1 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
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