Question

Consider the Ap 6,10,14,.. what is the sum of the 1st n consecutive terms of the above sequence How many consecutive terms from the beginning should be added together to get 240 Can the sum of the 1st few consecutive terms became 250 why?

Answer 1

Step-by-step explanation:

the AP is 6, 10, 14 ........

a = 6

d = 4

(a2 - a1 = a3 - a2;

10 - 6 = 14 - 10;

4=4)

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

$= \frac{n}{2} (2 \times 6 + (n - 1)4) \\ = \frac{n}{2} (12 + 4n - 4) \\ = \frac{n}{2} (8 + 4n) \\ = \frac{n}{2} \times 2(4 + 2n) \\ = n(4 + 2n) \\ = 4n + 2 {n}^{2} \\ = 2( {n}^{2} + 2n)$

number of terms required to get sum of 240

$240 = 2( {n}^{2} + 2n) \\ {n}^{2} + 2n = \frac{240}{2} \\ {n}^{2} + 2n = 120 \\ {n}^{2} + 2n - 120 = 0 \\ {n}^{2} + 12n - 10n - 120 = 0 \\ n(n + 12) - 10(n + 12) = 0 \\ (n + 12)(n - 10) = 0 \\ n + 12 = 0 \: \: \: \: \: n - 10 = 0 \\ n = - 12 \: \: \: \: \: \: \: \: \: \: n = 10$

we take the positive value of n

therefore number of terms to get sum of 240 is 10

hope you get your answer