Given series is in AP with first term as a = -7 and common difference d = 6 .
nth term of AP is given by Tn = a+(n-1)d
therefore 83 = -7 +(n-1)6
solving we get n = 16
hence given series is sum of first 16 term of an AP there fore we can write it as
question 2: 1-4+16 -64+...----
GIven series is in GP with a =1 and r = -4
sum of infinite series of a GP is given by : a/(1-r)
hence we can denote it as
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