in a.p.. a16=p, then find the sum of first 31 terms
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Step-by-step explanation:
nth term of AP =a+(n-1) d {equation 1}
a16=p (given)
a+15d=p {equation 2 from 1}
sum of first n terms in AP
Sn=n/2(2a+(n-1) d)
S31=31/2(2a+(31-1)d
31/2×2(a+15d)=31(a+15d)
a+15d=p {from equation 2}
S31=31p is your answer.
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