Question

Calculate the sum of first $30$ terms of the H.P.$-2,-5,-8,-11\ldots$

Answer 1

From the question given we have been asked to calculate the sum of first $30$ terms of the H.P.$-2,-5,-8,-11\ldots$. as we know that the reciprocal of arithmetic progression is harmonic progression. The formula for sum of nth terms in arithmetic progression is ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$, where “n” is number of terms, “a” is first term and “d” is common difference, after getting this sum. The reciprocal of that sum is the required sum of terms of the H.P.

Complete step by step solution:
from the question given we have to calculate the sum of first $30$ terms of the H.P.
$\Rightarrow -2,-5,-8,-11\ldots$
Here the first term is
$\Rightarrow a=-2$
And the common difference is
$\Rightarrow d=-3$
The formula for sum of nth terms in arithmetic progression is
$\Rightarrow {{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$
We need the sum of first $30$ terms that means
Here
$\Rightarrow n=30$
Now, the sum of first $30$ terms using the formula is,
$\Rightarrow {{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$
Now, we have to substitute the values in their respective positions,
By substituting the values, we will get,
$\Rightarrow {{S}_{30}}=\dfrac{30}{2}\left[ 2\times -2+\left( 30-1 \right)-3 \right]$
By simplifying further, we will get,
$\Rightarrow {{S}_{30}}=15\left[ -4+\left( 29 \right)-3 \right]$
By simplifying further, we will get,
$\Rightarrow {{S}_{30}}=15\left[ -4-87 \right]$
$\Rightarrow {{S}_{30}}=-1365$
Therefore, this is the answer for the sum of first $30$ terms in arithmetic progression but we need sum of first $30$ terms in H.P.
As we know that the reciprocal of arithmetic progression is harmonic progression.
Therefore, the reciprocal of this sum is the required answer.
$\Rightarrow {{S}_{30}}=\dfrac{-1}{1365}$.

Therefore, the sum of the first $30$ terms of the H.P. $-2,-5,-8,-11\ldots$ is ${{S}_{30}}=\dfrac{-1}{1365}$.

Note: Students should recall all the concepts and formula of progression before doing this question, student should also know about the G.P that is geometric progression, the formula for the sum of n terms of G.P is ${{S}_{n}}=\dfrac{a\left( {{r}^{n}}-1 \right)}{\left( r-1 \right)}$ where “n” is number of terms, “a” is first term and “r” is common ratio and $r\ne 1$, along with this the sum of infinite geometric progression is ${{S}_{n}}=\dfrac{a}{1-r}$;$-1 < r < 1$