Given: an = 1 – 4n Taking n = 1, a1 = 1 – 4(1) = 1 – 4 = -3 Taking n = 2, a2 = 1 – 4(2) = 1 – 8 = -7 Taking n = 3, a3 = 1 – 4(3) = 1 – 12 = -11 Therefore the series is -3, -7, -11, … So, a = -3, d = a2 – a1 = -7 – (-3) = -7 + 3 = -4 Now, we have to find the sum of the first 25 terms of the AP ⇒ S25 = 25 × (-51) ⇒ S25 = -1275 Hence, the sum of 25 terms is -1275.Read more on Sarthaks.com – https://www.sarthaks.com/945171/find-the-sum-of-first-25-terms-of-an-a-p-whose-nth-term-is-1-4nI hope it will help you?, ??
An=1-4n ,A1=1-4(1)=-3 ,A2=1-4(2)=-7 ,D=A2 – A1= -7-(-3)=-4 ,Sn = n/2{2A+(n -1)D} ,S25= 25/2{2×-3)+(24×-4)} ,…….25/2{-6 + (-96)} ,…….25/2 ×( -102) ,…….25 × (- 51) ,…….-1275 .
Find 1 term and 25 term and solve byn/2(a+l)