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

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Hint: Use the formula of Arithmetic progression sequence for the nth terms that is \[{a_n} = a + \left( {n - 1} \right)d\] 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 \[{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\]. Calculate the sum of the AP, \[{S_n}\].

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

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

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

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

Now, calculate the value of ${S_n}$ by substituting $n = 23$, $a = 2$ and $a_n = 149$ in the expression for the sum of the Arithmetic progression \[{S_n} = \dfrac{n}{2}\left[ {a + {a_n}} \right]\].
${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 \[{S_{22}} = 1,661\].

Note: The general equation of the Arithmetic progression is \[a,a + d,a + 2d,a + 3d,...\], 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 \[{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\] and use the Arithmetic progression sequence for the nth terms that is \[{a_n} = a + \left( {n - 1} \right)d\].