An arithmetic progression has first term 11 and fourth te...
An arithmetic progression has first term 11 and fourth term 32. The sum of the first nine terms is
351
531
135
153
Correct answer is A
1st term a = 11, 4th term = 32
nth term = a + (n - 1)d
4th term = 11 = (4 - 1)d
= 11 + 3d
= 32
3d = 21
d = 7
sn = n(2a + (n - 1)d)
sn = \(\frac{9}{2}\)(2 \times 11) + (9 - 1)7
\(\frac{9}{2}\)(22 + 56) = \(\frac{9}{2}\) x 78
= 351