If the sum of 1st 7 terms of an AP is 49 and that of 17 terms is 289. Find the sum of 1st n terms
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sum of first n terms of an a.p os n square.
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S 7 = 7/2 [2a+(6)d]
49×2/7 = 2a+6d
7×2 = 2a+6d
14 = 2a+6d........(1)
s17= 17/2 (2a+16d)
289×2/17 = 2a+16d
17×2 = 2a + 16d
34 = 2a + 16d .........(2)
subtracting eq. (2) from (1) we will get,
20 = 10 d
so ,d = 2
putting d = 2 in eq. 1
14 = 2a + 6 × 2
14 = 2a + 12
14 - 12 = 2a
2a = 2
a = 1
sn= n/2 [2a+ (n-1)d]
=n/2[2(n)]
=n2
sn =n2
49×2/7 = 2a+6d
7×2 = 2a+6d
14 = 2a+6d........(1)
s17= 17/2 (2a+16d)
289×2/17 = 2a+16d
17×2 = 2a + 16d
34 = 2a + 16d .........(2)
subtracting eq. (2) from (1) we will get,
20 = 10 d
so ,d = 2
putting d = 2 in eq. 1
14 = 2a + 6 × 2
14 = 2a + 12
14 - 12 = 2a
2a = 2
a = 1
sn= n/2 [2a+ (n-1)d]
=n/2[2(n)]
=n2
sn =n2
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