Short Answer Questions 2 — Question 10

(a) Let no. of terms be n.

By formula,

Sum of A.P. = $\dfrac{n}{2}[2a + (n - 1)d]$

Substituting values we get :

$\Rightarrow 65 = \dfrac{n}{2}[2 \times 20 + (n - 1) \times (-3)] \\[1em] \Rightarrow 65 = \dfrac{n}{2}[40 - 3n + 3] \\[1em] \Rightarrow 65 = \dfrac{n}{2}[43 - 3n] \\[1em] \Rightarrow 65 \times 2 = n(43 - 3n) \\[1em] \Rightarrow 130 = 43n - 3n^2 \\[1em] \Rightarrow 3n^2 - 43n + 130 = 0 \\[1em] \Rightarrow 3n^2 - 30n - 13n + 130 = 0 \\[1em] \Rightarrow 3n(n - 10) - 13(n - 10) = 0 \\[1em] \Rightarrow (3n - 13)(n - 10) = 0 \\[1em] \Rightarrow 3n - 13 = 0 \text{ or } n - 10 = 0 \\[1em] \Rightarrow 3n = 13 \text{ or } n = 10 \\[1em] \Rightarrow n = \dfrac{13}{3} \text{ or } n = 10.$

Since, no. of terms cannot be in fraction.

∴ n = 10.

Hence, no. of terms = 10.

(b) By formula,

Last term (l) = a + (n - 1)d

= 20 + (10 - 1) × (-3)

= 20 + 9 × (-3)

= 20 - 27

= -7.

Hence, last term = -7.