In the sum of first 7terms of an ap is 49 and that of it first 17terms is 289.find the sum of first n terms of the ap
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hello dude
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
so sum of first natural number = n2
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
so sum of first natural number = n2
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