Question

the 20th term of an AP exceeds it's 15th term by 15. find the common difference​

Answer 1

$\large{\underline{\bf{\purple{Given:-}}}}$

• ✦ 20th term exceeds 15th term by 15.

$\large{\underline{\bf{\purple{To\:Find:-}}}}$

• ✦ we need to find common difference.

$\huge{\underline{\bf{\red{Solution:-}}}}$

We know that :-

$\longmapsto \rm\:\:a_n= a+(n-1)d$

• $\rm\:a_{20}=a+19d\:$
• $\rm\:a_{15}=a+14d\:$

Now,

According to question:-

$\longmapsto \rm\:a_{20}-a _{15}=15$

$\longmapsto \rm (a + 19d) - (a + 14d) = 15 \\ \\ \longmapsto \rm \:{ \cancel{ a}} + 19d - { \cancel{a}} - 14d = 15 \\ \\ \longmapsto \rm \: 19d - 14d = 15 \\ \\ \longmapsto \rm5d = 15 \\ \\\longmapsto \rm \: \: d = \cancel\frac{15}{5} \\ \\\longmapsto \bf \: d = 3$

Hence ,

common difference = 3

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Answer 2

$\huge\red{\underline{\underline{\pink{Ans}\red{wer:-}}}}$

$\sf{The \ common \ difference \ is \ 3.}$

$\sf\orange{Given:}$

➡$\sf{20^{th} \ term \ of \ A.P. \ exceeds \ it's \ }$

$\sf{15^{th} \ term \ by \ 15.}$

$\sf\pink{To \ find:}$

$\sf{Common \ difference.}$

$\huge\sf\green{\underline{\underline{Solution:}}}$

$\sf{tn=a+(n-1)d...formula}$

$\sf{\therefore{=>t20=a+19d}}$

$\sf{\therefore{=>t15=a+14d}}$

$\sf{According \ to \ given \ condition}$

$\sf{=>t20-t15=15}$

$\sf{=>a+19d-(a+14d)=15}$

$\sf{=>a+19d-a-14d=15}$

$\sf{=>5d=15}$

$\sf{=>d=\frac{15}{5}}$

$\sf{=>d=3}$

$\sf\purple{\tt{\therefore{The \ common \ difference \ is \ 3.}}}$