Let S =4 · 9 + 7 · 14 + · · · + 28 · 49
every term in above sequence consists of two factors
i) first factors in the series are 4,7,10,. . . . .28 it is an arithmetic progression
first term = 4; common difference = 7 -4 =3
n th term = 4 + ( n- 1) 3 [ nth term = a + (n -1)d ]
= 4 + 3n - 3
=3n +1 -------------------------(1)
n th term = 28 (Given)
3n + 1 = 28
3n = 28 - 1
3n = 27
n = 27 / 3
n = 9 --------------(2)
ii) second factors in the series are 9,14 ,19 ,...... ,49 are in arithmetic progression
first term = 9; common difference = 14 -9 = 5
n th term = 9 + (n-1)5
= 9 + 5n - 5
= 5n +4 -----------------------(3)
n th term in the series = (1) *(3)
= (3n + 1) (5n + 1)
= 15n^2 + 3n + 5n + 1
= 15n^2 + 8n + 1
sum of the terms = sigma (15n^2 + 8n + 1)
=15*sigma n^2 + 8 *sigma *n + sigma 1
= 15 n(n+1)(2n+1)/6 + 8*n(n+1)/2 +n { sigma n^2 = n(n+1) (2n+1)/6 ; sigma n = n(n+1) /2 }
= 5/2 *n(n+1)(2n+1) +4* n(n+1) + n ----------------(4)
S = sum of 9 terms [ n = 9 from (2) ]
= 5/2 * 9(9+1)(2*9 +1) + 4 * 9 (9+1) + 9 [ substituting n=9 in (4) ]
= 5/2 * 9 * 10 *19 + 4 * 9 *10 +9
= 4275 + 360 + 9
= 4644