How Many Terms of the A.P. 54, 51, 48, has the Sum 513? Explain the Double Answer.

AP (Arithmetic Progression)

A sequence of numbers is said to be in arithmetic progression (AP) if there is a constant difference between any two successive integers in that sequence. Arithmetic Sequence is another name for AP in common usage.

Some formulas of Arithmetic Progression:

• Sum of the first n terms of an AP.

S, sum = S_{n} = \frac{n}{2}[2a+(n-1)d]

• Value of the nth term of AP.

n th term of AP, A_{n}=a+(n-1)d

Here,

a = first term

b = second term

n = number of terms

D =common difference = b - a

Given:

Arithmetic Progression, AP= 54, 51, 48, ……

The first term, a = 54

Second term = 51

Required Sum = 513

Solution:

Common difference, d = Second term - first term =51-54=-3

Required sum = 513

As the required sum is 513. So, substituting the given values to find the value of n, i,e., number of terms.

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

\Rightarrow 513 = \frac{n}{2}[2.54+(n-1)(-3)]

\Rightarrow 1026 = n (108 - 3n +3)

\Rightarrow 1026 = n (111 - 3n)

\Rightarrow 1026 = 111n - 3n^{2}

\Rightarrow 3n^{2}-111n + 1026 = 0

\Rightarrow 3n^{2} -57n - 54n + 1026 = 0

\Rightarrow 3n (n - 19) - 54 (n - 19) =0

\Rightarrow (3n - 54) (n - 19) = 0

\Rightarrow n=\frac{54}{3}=18 or n = 19.

Hence, the number of terms of the. AP 54, 51, 48, ...... so that their sum is 513 is 18 or 19.

Formula for nth term of an AP, A_{n}=a+(n-1)d

Where, n stands for the number of terms, a stands for first term and d stands for common difference

18th term, A_{18} = 54 +(18 - 1)(-3)=54 - 51 = 3

19th term, A_{19} = 54 +(19 - 1)(-3)=54 - 54 = 0

As the AP is in decreasing form. Where the 18th term is 3 and the 19th term is 0. So, the sum till the 18th term will be the same as the sum till the 19th term. As the number remains the same when added to 0.