Question

If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms

Answer 1

Let first term of AP = a

common difference = d

number of terms = n

Given that sum of first 7 term of AP = 49

=> (n/2)*{2a + (n-1)*d} = 49

=> (7/2)*{2a + (7-1)*d} = 49    

=> (7/2)*{2a + 6d} = 49

=> (1/2)*{2a + 6d} = 7

=> (2/2)*{a + 3d} = 7

=> a + 3d = 7...............1

Again given sum of first 17 term of AP = 289

=> (17/2)*{2a + (17-1)*d} = 289

=> (1/2)*{2a + 16*d} = 17          (when 17 and 289 is divided by 17)

=> (2/2)*{a + 8d} = 17

=> a + 8d = 17 ...............2

After solving equation 1 and 2, we get

a = 1 and d = 2

Now sum of n terms of AP = (n/2)*{2a + (n-1)*d}

=> (n/2)* {2*1 + (n-1)*2}

=> (n/2)* (2 + 2n - 2)

=> (n/2)*2n

=>  n2

Answer 2

S7=7/2(2a+(7-1)d)
49=7(a+3d)
7=a+3d....(1)
S17=17/2(2a+16d)
289=17(a+8d)
17=a+8d.....(2)
on subtracting (1)&(2),we get
5d=10
d=2
putting d=2,in equ. (1),then we get
a=1
then you solve it easily