Question

find the sum of first 52 term of an AP whose second and third term are 14 and 18 respectively​

Answer 1

a2 =14

a3=18

a+d =14

a+2d=18

now subtract eq

a+d=14

a+2d=18

we get -d = -4

d =4

a+d=14

a +4=14

a=10

S = n/2+(2a+(n-1)d

S =52/2+(2×10+(52-1)×4

S = 26+(20+51×4)

S =26+(20+204)

S=26+224

S = 250

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