Class 10

Math

All topics

Arithmetic Progressions

In an AP, it is given that $S_{5}+S_{7}=167$ and $S_{10}=235$, then find the AP, where $S_{n}$ denotes the sum of its first $n$ terms.

Let $a$ be the first term and $d$ be the common difference of the AP.

$S_{5}=2n [2a+(n−1)d]$

$=25 [2a+(5−1)d]$

$=25 [2a+4d]$

Similarly,

$S_{7}=27 [2a+6d]$

and $S_{10}=210 [2a+9d]=5(2a+9d)$

Now, $25 (2a+4d)+27 (2a+6d)=167$

$5a+10d+7a+21d=167$

$12+31d=167$ .........(i)

and $5(2a+9d)=235$

$2a+9d=47$ .......(ii)

Multiply (i) by (i) and (ii) by $6$,

$12a+31d=167$

$12a+54d=282$

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$−23d=−115⇒d=−23−115 =5$

From (i), $12a+31×5=167$

$12a+155=16⇒12a=167−155$

$12a=12⇒a=1212 =1$

Which implies: $a=1$ and $d=5$

Therefore, the AP is $1,6,11,16,....$.