Question

In the A.P. 7, 14, 21,…. How many terms are to be considered for getting sum 5740?
A.40
B.50
C.14
D. 51

Answer 1

Hint: First check weather given series is in A.P. are not. We can see that the common difference is 7. So it in A.P. we know the formula for sum of n terms in A.P. is \[{S_n} = \dfrac{n}{2} \left[ {2a + (n - 1)d} \right] \] . Where \[a \] is the first term in the given series, \[n \] is the number of terms in a given series, \[d \] is the common difference. Here \[{S_n} \] , \[a \] and \[d \] is given. We need to find \[n \] .

Complete step-by-step answer:
We have, \[{S_n} = \dfrac{n}{2} \left[ {2a + (n - 1)d} \right] \]
Here, \[{S_n} = 5740 \] , \[a = 7 \] , \[d = 7 \] and we need to find the value of \[n \] .
Substituting in above formula we get,
 \[ \Rightarrow 5740 = \dfrac{n}{2} \left[ {2 \times 7 + (n - 1)7} \right] \]
 (Using simple multiplication, we get)
 \[ \Rightarrow 5740 = \dfrac{n}{2} \left[ {14 + 7n - 7} \right] \]
 (Subtracting, we get)
 \[ \Rightarrow 5740 = \dfrac{n}{2} \left[ {7 + 7n} \right] \]
(Taking \[7 \] as common, we get)
 \[ \Rightarrow 5740 = \dfrac{n}{2} \times 7 \left[ {1 + n} \right] \]
(Multiply by 2 on both the sides, we get)
 \[ \Rightarrow 5740 \times 2 = 7n \left[ {1 + n} \right] \]
(Divided by 7 on both the sides, we get)
 \[ \Rightarrow \dfrac{{11480}}{7} = n \left[ {1 + n} \right] \]
 \[ \Rightarrow 1640 = {n^2} + n \]
Rearranging the above equation, we get a polynomial of degree 2. Meaning we get two roots,
 \[ \Rightarrow {n^2} + n - 1640 = 0 \]
Factorizing this we get,
 \[ \Rightarrow (n + 41)(n - 40) = 0 \]
 \[ \Rightarrow n = - 41 \] and \[n = 40 \]
 \[n \] Cannot be negative.
Hence, \[n = 40 \]
So, the correct answer is “Option A”.

Note: After solving the polynomial we choose the positive value because \[n \] is a number of terms in the given series. Which cannot be negative and zero. So we choose only the positive value which we have obtained \[n = 40 \] .Meaning that we need to consider 40 terms in A.P. so that we can obtain the sum 5740. We can verify this using the above formula for the sum of \[n \] terms in A.P. and substituting the value of \[n \] we get the sum 5740.