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If the 5th term of an A.P. is 31 and 25th term is 140 more than the 5th term, find the A.P.
Given:
The 5th term of an A.P. is 31 and 25th term is 140 more than the 5th term.
To do:
We have to find the A.P.
Solution:
Let the required A.P. be $a, a+d, a+2d, ......$
Here,
$a_1=a, a_2=a+d$ and Common difference $=a_2-a_1=a+d-a=d$
We know that,
$a_n=a+(n-1)d$
Therefore,
$a_{5}=a+(5-1)d$
$31=a+4d$
$a=31-4d$.....(i)
$a_{25}=a+(25-1)d$
$=a+24d$
According to the question,
$a_{25}=a_5+140$
$a+24d=31+140$
$31-4d+24d=171$
$20d=171-31$
$20d=140$
$d=\frac{140}{20}$
$d=7$
This implies,
$a=31-4(7)$
$=31-28$
$=3$
Therefore,
$a_1=3$
$a_2=a+d=3+7=10$
$a_3=a+2d=3+2(7)=3+14=17$
$a_4=a+3d=3+3(7)=3+21=24$
Hence, the required A.P. is $3, 10, 17, 24,......$
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