Let S = 3 + 7 + 14 + 24 + 37 +……….

By Shifting each term by one, we get,

S = 3 + 7 + 14 + 24 + 37 + ……+ nth term……….(1)

S = 3 + 7 + 14 + 24 +…….. + (n – 1)th term + nth term …(2)

Substracting equation 2 from equation 1 we get,

0 = 3 + 4 + 7 + 10 + 13 +……+ (nth term – (n – 1)th term) - nth term

Nth term = 3 + 4 + 7 + 10+…nth term – (n – 1)th term

We can see that 3, 4, 7,…is an A.P with first term = 3 and common difference = 3

Sum of this A.P

Therefore,

S

(n – 1)th term = a + (n – 2)d

(n – 1)th term = 3 + (n – 2)3

(n – 1)th term = 3n – 3

Therefore,

S

S