. If the sum of first seven terms of an AP is 49 and that of seventeen terms is 289. Find the sum of
first n terms
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Sn = n/2 (2a+(n-1)d)
sum of first 7 terms = 49
S7= 7/2 (2a+(7-1)d)
49=7/2 (2a+(6)d)
49×2/7=(2a+6d)
14 = 2a+6d
14-6d = 2a
14-6d/2 = a
a=7-3d.............1
sum of first seventeen terms = 289
S17 = 17/2 (2a+(17-1)d)
289 = 17/2 (2a+(16)d)
289×2/7 = (2a+16d)
34 = 2a+16d
34-16d = 2a
34-16d/2 = a
a=17-8d...............2
from (1) and (2)
7-3d = 17-8d
8d-3d = 17-7
5d = 10
d= 10/5
d=2
a = 7-3d
a = 7-3×2
a = 7-6
a = 1
Sn = n/2 (2a+ (n-1) d)
=n/2 (2×1 + (n-1) 2)
=n/2 (2 + (n-1) 2)
=n/2(2 + 2n - 2)
=n/2(0 + 2n)
=n/2 × 2n
=n×n
=n2
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