If the sum of 7 terms of an AP is 49 and that of 17 terms is 289. Find the sum of n terms...?
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Given that,
S7 = 49
S17 = 289
S7= 7/2 [2a + (n - 1)d]
S7 = 7/2 [2a + (7 - 1)d]
49 = 7/2 [2a + 16d]
7 = (a + 3d)
a + 3d = 7 ... (i)
Similarly,
S17 = 17/2 [2a + (17 - 1)d]
289 = 17/2 (2a + 16d)
17 = (a + 8d)
a + 8d = 17 ... (ii)
Subtracting equation (i) from equation (ii),
5d = 10
d = 2
From equation (i),
a + 3(2) = 7
a + 6 = 7
a = 1
Sn = n/2 [2a + (n - 1)d]
= n/2 [2(1) + (n - 1) × 2]
= n/2 (2 + 2n - 2)
= n/2 (2n)
= n2
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hope this helps.
shanasanu274:
thanks a lot
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