If the sum of first 7 terms of an ap is 49 and that of 17 terms is 289 find the 20th term

7 square=49

17 square=289

20 square=20*20=400

7th term. =a+6d

so, a+6d=49 (1)

17th term. =a+16d

so, a+16d=289 (2)

now subtracting eqn 1 from 2.

a+16d=289

a+6d=49

_ _ _

0+10d=240

thus, 10d=240

d=24( 3)

putting eqn 3in eqn 1

a+6×24=49

a=49-144

a=-95

now, 20th term= a+19d=-95+19×24

=-95+456

=361

Let the first term of the AP be a and the common difference be d.

Note: The sum of n terms of an AP is given by;

S(n) = (n/2){ 2a + (n-1)d }

Now,

It is given that, the sum of first 7 terms of the AP is 49.

=> S(n) = (n/2){ 2a + (n-1)d }

=> S(7) = (7/2){ 2a + (7-1)d }

=> 49 = (7/2){ 2a + 6d }

=> 49/7 = (2a + 6d)/2

=> 7 = a + 3d

=> a = 7 - 3d (1)

Also,

It is given that, the sum of first 17 terms of the AP is 289.

=> S(n) = (n/2){ 2a + (n-1)d }

=> S(17) = (17/2){ 2a + (17-1)d }

=> 289 = (17/2){ 2a + 16d }

=> 289/17 = (2a + 16d)/2

=> 17 = a + 8d

=> a = 17 - 8d (2)

From eq-(1) and (2) , we get;

=> 7 - 3d = 17 - 8d

=> 8d - 3d = 17 - 7

=> 5d = 10

=> d = 10/2

=> d = 5

Now, putting d = 5 in eq-(1) , we get;

=> a = 7 - 3d

=> a = 7 - 3•5

=> a = 7 - 15

=> a = - 8

Also, we know that the nth term of an AP is given by;

T(n) = a + (n-1)d

Thus,

T(20) = a + (20-1)d

= a + 19d

= - 8 + 19•5

= - 8 + 95

= 87

Hence, the 20th term of the AP is 87.

Step-by-step explanation:

BY the formulae

a+6d =49. (1)

and

a+16d=289 (2)

then, by elimination

10d = 240

d = 24

put the value of d in equation no 1 the we get

a+6(24)= 49

a+144. =49

a. =49-144

a. =-95

now we can solve for the value of the 20th term

20 th term = a+19d

which gives

=-95+19(24)

=-95+456

=361

Hence the answer is 361

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