Question

find the sum of the first 51 terms of an a.p. whose second and third terms are 14 and 18

Answer 1

second term = 14
a2 = 14
a + d = 14
a = 14 - d................(1)

third term = 18
a3 = 18
a + 2d = 18
14 - d + 2d = 18......(using 1)
14 + d = 18
d = 4

put this in equation(1)
a = 14 - 4
a = 10

sum of 51 terms , Sn = n/2 { 2a + ( n-1)d}
S51 = 51/2 { 2× 10 + ( 51 -1) × 4}
= 51/2 {20 + 200}
= 51/2 × 220
= 51 × 110
= 5610 is the answer.

Answer 2

d=18-14=4
Second term =14
Third term = 18
T2=a+(n-1)*d
14=a+(2-1)*4
14=a+1*4
14=a+4
14-4=a
10=a
S51=n/2(2a+(n-1)d)
=51/2(2*10+(51-1)×4)
=51/2(20+(50*4)
= 51/2(20+200)
=(51/2)*220
=51*110
=5610

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