Question

What is this sum equal to
$\sum\limits_{p=1}^{25}{\left( 12p+7 \right)}$
[a] 4200
[b] 4075
[c] 4100
[d] 4050

Answer 1

Hint: Use the property \[\sum{\left( a+b \right)=\sum{a}+\sum{b}}\] and $\sum{ka}=k\sum{a}$ where k is a constant. Use \[\sum\limits_{r=1}^{n}{r}=\dfrac{n\left( n+1 \right)}{2}\] and $\sum\limits_{r=1}^{n}{1}=n$. Alternatively, express this sum as ${{S}_{n}}$ of some A.P and use ${{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$

Complete step-by-step answer:

Let $S=\sum\limits_{p=1}^{25}{\left( 12p+7 \right)}$
We know that \[\sum{\left( a+b \right)=\sum{a}+\sum{b}}\]
Using the above property, we get
$S=\sum\limits_{p=1}^{25}{12p}+\sum\limits_{p=1}^{25}{7}$
We know that $\sum{ka}=k\sum{a}$ where k is a constant
Using the above property, we get
$S=12\sum\limits_{p=1}^{25}{p}+7\sum\limits_{p=1}^{25}{1}$
We know that \[\sum\limits_{r=1}^{n}{r}=\dfrac{n\left( n+1 \right)}{2}\] and $\sum\limits_{r=1}^{n}{1}=n$
Using the above identities for n = 25, we get
$\begin{align}
  & S=12\times \dfrac{25\times 26}{2}+7\times 25 \\
 & =4075 \\
\end{align}$
Hence we have S = 4075
Hence option [b] is correct.

Note: [1] Alternate Solution.
Expressing the individual terms of the series as a sequence we get the following sequence
19,31,43,55,… which is an A.P with a = 19 and d = 12.
Using ${{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$
Put n = 25 , a= 13 and d = 12 we get
${{S}_{n}}=\dfrac{25}{2}\left( 2\times 19+24\times 12 \right)=4075$
which is the same as obtained above.
[2] Any series in which nth term is linear is a series of numbers in arithmetic progression.
Consider the following series $\sum\limits_{r=1}^{n}{a+br}$.
Our claim is that the series is a series of numbers in arithmetic progression.
Let us write a few terms of this series
a+b, a+2b, a+3b, …
which clearly is an A.P with the first term as a+b and common difference = b.
[3] If for any series we have ${{S}_{n}}$ is quadratic with no constant term, then the series is a series of numbers in arithmetic progression.
Let ${{S}_{n}}=\sum\limits_{r=1}^{n}{{{a}_{r}}=A{{n}^{2}}+Bn}$ then $\left\{ {{a}_{n}} \right\}$ is an Arithmetic progression.
Proof:
$\begin{align}
  & {{a}_{n}}={{S}_{n}}-{{S}_{n-1}} \\
 & =A\left( {{n}^{2}}-{{\left( n-1 \right)}^{2}} \right)+B\left( n-\left( n-1 \right) \right) \\
 & =\left( 2n-1 \right)A+B \\
\end{align}$
Since ${{a}_{n}}$ is linear, from above we have ${{a}_{n}}$ is an A.P