An arithmetic progression has first term 11 and fourth term 32. The sum of the first nine terms is
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Correct Answer: Option A
Explanation:
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
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