Question

in an AP,6th terms half of the 4th term and 3rd term is 15 . how many terms are needed to give a sum of 66?​

Answer 1

Answer:

4 terms, or 11 terms

Step-by-step explanation:

$a_{n} = a_{1} + (n-1)d$

$S_{n} =\frac{n}{2}[2a_{1} +(n-1)d]$

~~~~~~~~

$a_{3} = a_{1} +2d$ = 15

$a_{4} = a_{1} +3d=2(a_{1} +5d)$ ⇒ $a_{1} = -7d$  

- 7d + 2d = 15 ⇒ d = - 3  

$a_{1}$ = 21

(n/2)[2×21 + (- 3)(n - 1)] = 66

n[42 - 3n + 3] = 132

3n² - 45n + 132 = 0 ⇔ n² - 15n + 44 = 0

$n_{1}$ = 4

$n_{2}$ = 11

$S_{4} =$ 21 + 18 + 15 + 12 = 66

$S_{11} =$ 21 + 18 + 15 + 12 + [9 + 6 + 3 + 0  + (- 3) + (- 6) + (- 9)] = 66 + 0 = 66