Question

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

Answer 1

Answer:

S7 = 7(a+l)/2

49= 7(a+l)/2

a+l = 14

2a+6d = 14

a+3d =7

a4 = 7

similarly,

S17 = 17(a+a17)/2

2a+16d = 34

a9 = 17

solving for d we get,

5d = 10

d = 2

a+3(2) = 7

a = 1

therefore the series is the series of all odd natural numbers.

Sn = n(2+(n-1)2)/2

Sn = n(2+2n-2)/2

Sn = n×2n/2

Sn = n^2

Hope this helps

Answer 2

Step-by-step explanation:

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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