Question

How many terms of an A.P. 24, 21, 18,…. Must be taken so that their sum is

78.​

Answer 1

Answer:

13 terms

Step-by-step explanation:

Given :

a = first term = 24

d = common difference = - 3

Sum of the terms = 78

To find:

Number of terms so that the sum of the following Arithmetic progression will be 78

Sum of n terms of an A.P =

(n/2) (2a+(n-1)d)

Substituting the values,

$\frac{n}{2} \times (2 \times 24 + (n - 1) - 3 = 78</p><p>$

78×2 = n (2×24+(n-1)-3)

156 = n (48-3n+3)

156 = n (51-3n)

156 = 51n - 3n²

-3n²+51n-156

-3n²+39n+12n-156

-3(n-13)+12(n-13)

n-13=0

n = 13

13 terms shall be considered to get the sum as 78

Answer 2

Given ,

First term (a) = 24

Common difference (d) = -3

Sum of n terms (Sn) = 78

We know that , the sum of First n terms of an AP is given by

$\star \: \: \sf S_{n} = \frac{n}{2} (2a + (n - 1)d$

Thus ,

$\sf \Rightarrow 78 = \frac{n}{2} (2 \times 24 + (n - 1) \times (- 3) \\ \\ \sf \Rightarrow 156 = n \times (48 - 3n + 3) \\ \\\sf \Rightarrow 156 = 48n - 3 {n}^{2} + 3n \\ \\ \sf \Rightarrow 3 {n}^{2} - 51n + 156 = 0 \\ \\ \sf \Rightarrow 3 {n}^{2} - 39n - 12n + 156 = 0 \\ \\\sf \Rightarrow 3n(n - 13) - 12(n - 13) = 0 \\ \\\sf \Rightarrow (3n - 12)(n - 13) = 0 \\ \\\sf \Rightarrow 3n - 12 = 0 \: \: or \: \: n - 13 = 0 \\ \\\sf \Rightarrow n = 4 \: \: or \: \: n = 13$

$\therefore \sf \bold{ \underline{The \: sum \: of \: first \: 4 \: or \: 13 \: terms \: of \: AP \: is \: 78}}$