Question

Let ${a_1},{a_2},{a_3},{\text{ - - - - - - - }}{{\text{a}}_{49}}$ be in A.P. such that $\sum\limits_{k = 0}^{12} {{a_{4k + 1}}} = 416$ and ${a_9} + {a_{43}} = 66$. If ${a_1}^2 + {a_2}^2 + {\text{ - - - - - - - - + }}{a_{17}}^2 = 140m$, then $m$ is equal to
A) $34$
B) $33$
C) 66
D) 68

Answer 1

Hint: Here given ${a_1},{a_2},{a_3},{\text{ - - - - - - - }}{{\text{a}}_{49}}$are in A.P.
Let ${a_1} = a$ and common difference$ = d$
$\sum\limits_{k = 0}^{12} {{a_{4k + 1}}} = 416$ that means ${a_1} + {a_5} + {a_9} + {\text{ - - - - - - - - - + }}{{\text{a}}_{49}} = 416$
So, we can put ${a_1} = a$, ${a_5} = a + 4d$, ${a_9} = a + 8d$ and so on. In this way we can find a relation between $a$ and $d$.

Complete step-by-step answer:
So, here according to the question, it is given that ${a_1},{a_2},{a_3},{\text{ - - - - - - - }}{{\text{a}}_{49}}$are in A.P.
So, let the first term of A.P. that is ${a_1}$ be $a$ and the common difference be $d$.
Then the ${n^{th}}$term of A.P. is given by
${T_n} = a + \left( {n - 1} \right)d$
So, ${a_5} = a + 4d$
And here it is also given that $\sum\limits_{k = 0}^{12} {{a_{4k + 1}}} = 416$. So, this means if we put $k = 0{\text{ , }}k = 1$ and sum it up, we will get ${a_1} + {a_5} + {a_9} + {a_{13}}{\text{ + }}{{\text{a}}_{17}}{\text{ + - - - - - - - - - + }}{{\text{a}}_{49}} = 416$
Now we can write ${a_5} = a + 4d$, ${a_9} = a + 8d$ and so on.
$
  a + a + 4d + a + 8d + a + 12d + {\text{ - - - - - - - - - }} + a + 48d = 416 \\
  a\left( {1 + 1 + {\text{ - - - - - }}13{\text{ times}}} \right) + 4d\left( {1 + 2 + 3 + {\text{ - - - - - + }}12} \right) = 416 \\
  13a + 4d\left( {1 + 2 + 3 + {\text{ - - - - - + }}12} \right) = 416……………….{\text{ (1)}} \\
 $
And we know the formula of
$
  1 + 2 + 3 + 4 + {\text{ - - - - - - - }}n = \dfrac{{n\left( {n + 1} \right)}}{2} \\
  {\text{so, for }}n = 12, \\
  {\text{we get}} \\
  1 + 2 + 3 + {\text{ - - - - - 12 = }}\dfrac{{12 \times 13}}{2} = 78 \\
 $
So, we get
$
\Rightarrow 13a + 4d\left( {78} \right) = 416 \\
\Rightarrow 13\left( {a + 24d} \right) = 416 \\
\Rightarrow a + 24d = 32…………………...{\text{ (2)}}
 $
Also, in question it is given that
$
  {a_9} + {a_{43}} = 66 \\
   \Rightarrow a + 8d + a + 42d = 66 \\
  2a + 50d = 66 \\
  a + 25d = 33…………………...{\text{ (3)}}
 $
Now subtract equation (2) from (3),
$
  \left( {a + 25d} \right) - \left( {a + 24d} \right) = 33 - 32 \\
  d = 1 \\
 $
So,
$
\Rightarrow a + 24d = 32 \\
 \Rightarrow a + 24 = 32 \\
\Rightarrow a = 8
 $
Here we got $a = 8$ and $d = 1$
So, we have ${a_1} = 8$, ${a_2} = 8 + d$${a_3} = 8 + 2d$ and so on.
${a_1} = 8,{\text{ }}{a_2} = 9,{\text{ }}{a_3} = 10,{\text{ }}{a_4} = 11{\text{ and so on}}..$${a_{17}} = a + 16d = 8 + 16 = 24$
And now we are given ${a_1}^2 + {a_2}^2 + {\text{ - - - - - - - - + }}{a_{17}}^2 = 140m$
We have to find the value of $m$
${a_1}^2 + {a_2}^2 + {\text{ - - - - - - - - + }}{a_{17}}^2 = 140m$
So, putting the values, we get
${8^2} + {9^2} + {\text{ - - - - - - - - - + 2}}{{\text{4}}^2} = 140m$
We know that
${1^2} + {2^2} + {3^2} + {\text{ - - - - - - - - - + }}{n^2} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}$
Now, if $n = 24$, we get
${1^2} + {2^2} + {3^2} + {\text{ - - - - - - + 2}}{{\text{4}}^2} = \dfrac{{24 \times 25 \times 49}}{6}....................{\text{ (4)}}$
Now, if $n = 7$, we get
${1^2} + {2^2} + {3^2} + {\text{ - - - - - - + }}{{\text{7}}^2} = \dfrac{{7 \times 8 \times 15}}{6}...................{\text{ (5)}}$
Now subtracting equation (5) by (4), we get
$
 \Rightarrow {8^2} + {9^2} + {10^2} + {\text{ - - - - - - - - + 2}}{{\text{4}}^2} = \dfrac{{24 \times 25 \times 49}}{6} - \dfrac{{7 \times 8 \times 15}}{6} \\
   = 4 \times 25 \times 49 - 7 \times 4 \times 5
 $
As we know ${8^2} + {9^2} + {\text{ - - - - - - - - - + 2}}{{\text{4}}^2} = 140m$
So,
$
\Rightarrow 140m = 4900 - 140 \\
\Rightarrow 140m = 4760 \\
 \Rightarrow m = \dfrac{{4760}}{{140}} = 34 \\
  m = 34
 $

So, option A is the correct answer.

Note: We must know the basic formulas that
$1 + 2 + 3 + 4 + {\text{ - - - - - - - }}n = \dfrac{{n\left( {n + 1} \right)}}{2}$
${1^2} + {2^2} + {3^2} + {\text{ - - - - - - - - - + }}{n^2} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}$
${1^3} + {2^3} + {3^3} + {\text{ - - - - - - - + }}{n^3} = {\left[ {\dfrac{{n\left( {n + 1} \right)}}{2}} \right]^2}$
And we know sum of A.P. is ${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$