Question

The sum of first 8 terms of an A.P is 100 and \[{S_{19}} = 551\]. Find \[a\] and \[d\].

Answer 1

Hint:
The sum of n terms of AP is the sum(addition) of first n terms of the arithmetic sequence. It is equal to n divided by 2 times the sum of twice the first term – ‘a’ and the product of the difference between second and first term-‘d’ also known as common difference, and (n-1), where n is the number of terms to be added. Use this relation to find two linear equations of two variables and solve them to get the values of a and d.

Complete step by step solution:
It is given that,
The sum of the first 8 terms of the A.P is = 100
The sum of the first 19 terms is = 551
Let The first term of the A.P be \[a\]and the common difference be \[d\].
The sum of n terms of AP is the sum(addition) of first n terms of the arithmetic sequence. It is equal to n divided by 2 times the sum of twice the first term – ‘a’ and the product of the difference between second and first term- ‘d’ also known as common difference, and (n-1), where n is the number of terms to be added.
Sum of n terms of AP = \[\dfrac{n}{2}[2a + (n - 1)d]\]
We have been given that,
\[{S_8} = 100\]
\[ \Rightarrow \dfrac{8}{2}[2a + (8 - 1)d] = 100\]
\[ \Rightarrow 4[2a + 7d] = 100\]
\[ \Rightarrow 2a + 7d = 25\]……… (i)
And, \[{S_{19}} = 551\]
\[ \Rightarrow \dfrac{{19}}{2}[2a + (19 - 1)d] = 551\]
\[ \Rightarrow 19[a + 9d] = 551\]
\[ \Rightarrow a + 9d = 29\]…………… (ii)
We will solve eq. (i) and eq. (ii) to get the values of a and d.
First, multiply eq. (ii) with 2. We will get,
\[2a + 18d = 58\]…………… (iii)
Now. Subtract eq. (i) from eq. (iii). So, We get,
\[2a + 18d - 2a - 7d = 58 - 25\]
\[ \Rightarrow 11d = 33\]
\[ \Rightarrow d = 3\]
Now, substitute \[d = 3\] in eq. (i). We will get the value of a.
\[2a + 7 \times 3 = 25\]
\[ \Rightarrow 2a + 21 = 25\]
\[ \Rightarrow 2a = 25 - 21\]
\[ \Rightarrow 2a = 4\]
\[ \Rightarrow a = 2\]

Hence, the required values are $a=2$ and $d=3$.

Note:
Once you find the value of a and d, you can form the A.P as follows.
The AP will be \[a, a + d, a + 2d, a + 3d,............, a + (n - 1)d\].
So, the AP is: 2, 5, 8, 11, 14 and so on.