Question

How many terms of the A.P: 24, 21, 18……............ must be taken so that their sum is 78?

Answer 1

Hint: In this question directly use the formula for sum of n terms of an A.P that is ${S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$, where n the required number of terms, a is the first term and d is the common difference between the terms.

Complete step-by-step answer:

The given A.P is
$24,21,18,...................$
So the first term (a) of the A.P is = 24.
And the common difference (d) of the A.P is
$ \Rightarrow d = \left( {21 - 24} \right) = \left( {18 - 21} \right) = - 3$
Now we have to find out the number of terms in the A.P if the sum of an A.P is 78.
As we know that the sum (Sn) of an A.P is given as
$ \Rightarrow {S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$
Where symbols have their usual meanings.
So substitute the values in the above equation we have,
$ \Rightarrow 78 = \dfrac{n}{2}\left( {2 \times 24 + \left( {n - 1} \right)\left( { - 3} \right)} \right)$
Now simplify the above equation we have,
$
   \Rightarrow 156 = n\left( {48 - 3n + 3} \right) \\
   \Rightarrow 156 = n\left( {51 - 3n} \right) \\
   \Rightarrow 3{n^2} - 51n + 156 = 0 \\
$
Now divide by 3 throughout we have,
$ \Rightarrow {n^2} - 17n + 52 = 0$
Now factorize this equation we have,
$ \Rightarrow {n^2} - 13n - 4n + 52 = 0$
$ \Rightarrow n\left( {n - 13} \right) - 4\left( {n - 13} \right) = 0$
$ \Rightarrow \left( {n - 13} \right)\left( {n - 4} \right) = 0$
$ \Rightarrow n = 4,13$
So the number of terms of the A.P so that their sum is 78 are 4 and 13.
So this is the required answer.

Note: A specific series is said to be A.P if and only if the difference between consecutive terms remains constant throughout. Similarly if we talk about G.P then the common ratio needs to be constant throughout the series. It is advised to learn basic series formulas as it helps save a lot of time.