Question

How do you find the ${10^{th}}$ partial sum of the arithmetic sequence $40,37,34,31,...$?

Answer 1

In this question, we have to find out the required value from the given particulars.
We need to first find out the common difference & the first term. By subtracting the first term from the second term we will get the common difference .Then putting all the values and the number of terms in the formula of the sum of nth partial sum of the arithmetic sequence, we can find out the required solution.

Formula used: Property of A.P.:
The nth term of the arithmetic sequence is
${a_n} = a + \left( {n - 1} \right)d$
The sum of nth partial sum of the arithmetic sequence is
${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$
Where,
a = first term of the sequence
d = common difference
n= number of terms

Complete step-by-step solution:
It is given the arithmetic sequence $40,37,34,31,...$.
We need to find the ${10^{th}}$ partial sum of the arithmetic sequence $40,37,34,31,...$.
a = the first term of the arithmetic sequence =$40$.
d = the common difference = second term – first term = $37 - 40 = - 3$.
n = number of terms = $10$.
The sum of ${10^{th}}$partial sum of the arithmetic sequence is
${S_{10}} = \dfrac{{10}}{2}\left[ {2 \times 40 + \left( {10 - 1} \right) \times \left( { - 3} \right)} \right]$
On simplification we get
Or,${S_{10}} = \dfrac{{10}}{2}\left[ {80 - 9 \times 3} \right]$
Let us divide the terms and we get
Or, ${S_{10}} = 5 \times \left[ {80 - 27} \right]$
On subtracting we get
Or,${S_{10}} = 5 \times 53$
Let us multiply we get
Or,${S_{10}} = 265$

Hence, the ${10^{th}}$ partial sum of the arithmetic sequence $40,37,34,31,...$ is $265$.

Note: An arithmetic progression is a sequence of numbers such that the difference of any two successive members is a constant.
In General we write an Arithmetic Sequence like this: \left\\{ {a,a + d,a + 2d,a + 3d....} \right\\} where a is the first term, and d is the difference between the terms (Called the “common difference”).