Q19 of 49 Page 1

The nth term of an A.P. is given by (-4n + 15). Find the sum of first 20 terms of this A.P.

nth term of an A.P(an) = -4n+15


∴ a1 = -4(1)+15 = 11


a2 = -4(2)+15 = 7


d = a2- a1 = 7-11 = -4


And, Sum of first n terms of AP is given by-


Sn = (n/2)[2a+(n-1)d]


= (n/2)[2×11 + (n-1)(-4)]


= (n/2)[22-4n+4]


= (n/2)(26-4n)


= n(13-2n)


∴ S20 = 20(13-2×20) = 20(13-40) = 20(-27) = -540


Thus, the Sum of first 20 terms of AP = -540


More from this chapter

All 49 →
17

Prove that the points A (0, -1), B (-2, 3), C (6, 7) and D (8, 3) are the vertices of a rectangle ABCD?

 18

Draw a triangle PQR in which QR = 6 cm, PQ = 5 cm and ∠ PQR = 60°. Then construct another triangle whose sides are (3/5) times the corresponding sides of Δ PQR?

 20

For what value of k, the roots of the quadratic equation kx(x-2√5)+10 = 0 are equal?

 28

Find the value of x for which the points (x, 1), (2, 1) and (4, 5) are collinear.