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Question

In an AP, given an=4,d=2,Sn=14, find n and a.

A
n=11 and a=3
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B
n=10 and a=3
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C
n=7 and a=8
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D
n=10 and a=3
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Solution

The correct option is C n=7 and a=8
As we know nth term, an=a+(n1)d
& Sum of first n terms, Sn=n2(2a+(n1)d), where a & d are the first term amd common difference of an AP.
Since, an=4a+(n1)d=4
a+(n1)2=4
a=62n ...(1)
Also, Sn=14
n2[2a+(n1)d]=14
n2[2(62n)+(n1)2]=14 (from (1))
n[124n+2n2]=28
n[n5]=14
n25n14=0
(n7)(n+2)=0
n=7,2
But, n=2 is not possible.
n=7
Putting n=7 in (1), we get
a=62×7=8

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