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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