How many term of ap 63, 60, 57 must be taken so that sum is 693?

1.	How many term of ap 63, 60, 57 must be taken so that sum is 693?
Answer» $\huge{\bf{\underline{\purple{Solution :}}}}$ GIVEN, First term (a) = 63 Common DIFFERENCE (d) = 60 - 63 = -3 To Find, 63 , 60 , 57 ....?.. = 693 Solution, ⇒Let the number of terms be n ∵ $\bold{s_{n} = 693}$ We know that : $\boxed{\bold{s_{n} = \frac{n}{2} [ 2a + (n - 1) d}}$ Here , a is the first term . n is the number of terms . d is the difference between two Consecutive terms $\implies \bold{693 = \frac{n}{2} [2 \times 63 + (n - 1) -3]}$ $\implies \bold{ 693 = \frac{n}{2} [126 - 3n + 3]}$ $\implies \bold{ 1386 = n(129 - 3n)}$ $\implies \bold{ 1386 = 129n - 3n}$ $\implies \bold{3n^{2} - 129n +1386 = 0}$ $\implies \bold{n^{2} - 43n + 462 = 0}$ $\implies \bold{n^{2} - 21n - 22n + 462 = 0}$ $\implies \bold{n(n - 21) - 22(n - 21) = 0}$ $\implies \bold{(n - 21) (n - 22) = 0}$ Therefore , $\boxed{\bold{ number \ of \ terms = 21 \ or \ 22}}$ $\rule{200}2$

Answer»

$\huge{\bf{\underline{\purple{Solution :}}}}$