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Question: Find the sum of the first 22 terms of an AP in which \[d = 7\] and \({22^{nd}}\) term is 149....

Find the sum of the first 22 terms of an AP in which d=7d = 7 and 22nd{22^{nd}} term is 149.

Explanation

Solution

Use the formula of Arithmetic progression sequence for the nth terms that is an=a+(n1)d{a_n} = a + \left( {n - 1} \right)d where, a initial term of the AP and d is the common difference of successive numbers. Calculate the value of a. We use the formula of the sum of n terms in Arithmetic progression that is Sn=n2[2a+(n1)d]{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]. Calculate the sum of the AP, Sn{S_n}.

Complete step by step solution:
Given data: The 22nd{22^{nd}} term that is given for an arithmetic progression is 149.
Common difference is d=7d = 7
Now, we know about the Arithmetic progression sequence for the nth terms is given by the following expression:
an=a+(n1)d{a_n} = a + \left( {n - 1} \right)d
Here, the first term of the arithmetic progression sequence is aa.
Now, calculate the value of aa. Substitute the value of d = 7,n = 22 and an=149{a_n} = 149 in an=a+(n1)d{a_n} = a + \left( {n - 1} \right)d.
149 = a + (22- 1)7
149 = a + 147
a = 149 - 147
= 2
Now, we know about the formula of the sum of n terms in Arithmetic progression is given by the following expression:

Sn=n2[2a+(n1)d]{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]

Simplify the above equation by substituting an=a+(n1)d{a_n} = a + \left( {n - 1} \right)d.

Sn=n2[a+an]{S_n} = \dfrac{n}{2}\left[ {a + {a_n}} \right]

Now, calculate the value of Sn{S_n} by substituting n=23n = 23, a=2a = 2 and an=149a_n = 149 in the expression for the sum of the Arithmetic progression Sn=n2[a+an]{S_n} = \dfrac{n}{2}\left[ {a + {a_n}} \right].
S22=222[2+149] =11[151] =1,661{S_{22}} = \dfrac{{22}}{2}\left[ {2 + 149} \right]\\\ = 11\left[ {151} \right]\\\ = 1,661
Hence, the sum of the first 22 terms of an Arithmetic progression is S22=1,661{S_{22}} = 1,661.

Note: The general equation of the Arithmetic progression is a,a+d,a+2d,a+3d,...a,a + d,a + 2d,a + 3d,..., where a is initial term of the AP and d is the common difference of successive numbers. Make sure use the formula of the sum of n terms in Arithmetic progression that is Sn=n2[2a+(n1)d]{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] and use the Arithmetic progression sequence for the nth terms that is an=a+(n1)d{a_n} = a + \left( {n - 1} \right)d.