Question

Solve the 14th question plzzz!!

Answer 1

Given that sum of first 7 terms = 49.

We know that sum of n terms sn = n/2(2a + (n - 1) * d)

                                                  s7 = 7/2(2a + (7 - 1) * d)

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

                                                  49 = 7/2 * 2(a + 3d)

                                                  49 = 7(a + 3d)

                                                   7 = (a + 3d)   ------------------ (1)




Given that sum of 17 terms of an AP = 289.

s17 = 17/2(2a + (17 - 1) * d

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

289 = 17/2 * 2(a + 8d)

289/17 = a + 8d

17 = a + 8d   --------------------------- (2)


On solving (1) & (2), we get

a + 8d = 17

a + 3d = 7

----------------

      5d = 10

        d = 2.


Substitute d = 2 in (2), we get

a + 8d = 17

a + 8(2) = 17

a + 16 = 17

a = 17 - 16

a = 1.


Now,

Sum of first n terms Sn = n/2(2a + (n - 1) * d)

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

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

                                       = n/2 * 2n

                                       = n^2.



Hope this helps!

Answer 2

Hi,

Please see the attached file !


Thanks