Question

24,21,18are in a.p find the sum of n terms .if the sum of n terms is 78​

Answer 1

Answer:

24, 21, 18 are in A.P.

Sn = 78

We have to find, n=?

Here, a = 24, d = -3

We know,

$sn = \frac{n}{2}(2a + (n - 1)d) \\ 78 = \frac{n}{2}(2 \times 24 + (n - 1)( - 3)) \\ 78 = \frac{n}{2}(48 - 3n + 3) \\ 78 \times 2 = n(51 - 3n) \\ 156 = 51n - 3 {n}^{2} \\ 3 {n}^{2} - 51n + 156 = 0$

Divide by 3

${n}^{2} - 17n + 52 = 0 \\ {n}^{2} - 4n - 13n + 52 = 0 \\ n(n - 4) - 13(n - 4) = 0 \\ (n - 4)(n - 13) = 0 \\ n - 4 = 0 \: \: \: \: \: or \: \: \: \: \: n - 13 = 0 \\ n = 4 \: \: \: \: \: or \: \: \: \: \: n = 13$

The number of terms is 4 or 13

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Answer 2

Answer:

HELLO

PLZ FOLLOW ME

24,21,18.. IS A.P

A = 24 , D= -3

S78 = ?

S78 = n/2 (2a + (n-1) d )

after solving these

UR ANSWER IS 4 ANF 13...

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