SSC CGL Model Paper 8

Question : 64 of 100

Marks: +1, -0

What is the sum of the first 12 terms of an arithmetic progression if the first term is 3 and last term is 47?

Solution:

T_n = a + (n - 1) d

⇒

47 =3 + (12 -1 ) ×d

⇒

44 =11d

∴

d =4

Hence,

S_n =n/2[2a+(n-1)d]

S_12 =12/2[2×3+(12-1)×4]

= 6 [6+44] =6 × 50 = 300

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