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Find the sum of the following arithmetic progression \[a + b,a - b,a - 3b......\]to \[22\]terms.

Last updated date: 20th Jun 2024
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Hint: Arithmetic progression or arithmetic sequence is a sequence of No. such that the difference between the consecutive terms constant difference here means the second minus the first for instance the sequence \[5,7,9,11,13,15,.......\] is an A.P with a common difference.
In initial term of an AP is a and common difference of successive members is d than the \[{n^{th}}\] term of the sequence (an) is given by
\[an = a,\alpha (n - 1)d\]
\[an = am + (n - m)d\]
A finite portion of an A.P is called a finite arithmetic progression and sometimes just called an A.P. the sum of a finite A.P is called an arithmetic series.
The behaviors of A.P. depend on the common difference d. If the common difference is.
Positive, then members will grow towards +ve infinity
–ve them member will grow towards –ve infinity

Complete step by step answer:

A.P is \[a + b,a - b............22\] terms
First team \[ = A = a + b\]
Common different \[d = a - b - a - b = - 2b\]
Then as we known
Sum of –h terms in an A.P. is
\[S = \dfrac{n}{2}\left[ {2A + (n - 1)d} \right]\]
Sum of 22 terms \[n( = 22)\]
\[{S_{22}} = \dfrac{{22}}{7}\left[ {2A + (22 - 1)d} \right]\]
\[ = 11\left[ {20(a + b) + 21( - 2b)} \right]\]
\[ = 11\left[ {2a + 2b - 42b} \right]\]
\[ = 11\left[ {2a - 40b} \right]\]
\[ = 11 \times 2\left[ {a - 20b} \right]\]
\[ = 22\left[ {a - 20b} \right]\]
Sum of 22 terms of a given A.P is \[22\left[ {a - 20b} \right]\]

Note: It will be very difficult to solve the question to sum the digits of an A.P. up to 22 terms. So, formula of sum of a given AP is
\[Sn = \dfrac{n}{2}\left[ {2a + (n - 1)d} \right]\]
Where \[n = No.\]of terms up to sum
\[a = \] first no
\[d = \] common difference
The best-case scenario for this question