Question

An A.P. consists of 37 terms, the sum of the three middle most terms is 225 and the sum of the last three terms is 429. Find the A.P. and the sum of all the 37 terms.

Answer 1

Hint First of all, let the first term of an A.P. is $a$ and let $d$ be the common difference of A.P. Then, we will use the given conditions to form equations in $a$ and $d$. Form the corresponding A.P. and then find the sum of first 37 terms of the A.P by using the direct formula.

Complete step-by-step answer:
Since, we are given A.P., let the first term of an A.P. is $a$ and let $d$ be the common difference of A.P.
We are given that the sum of the three middle most terms is 225.
We will now find the term which is in the middle.
Since, there are 37 terms, the middle most term of the A.P. is ${\left( {\dfrac{{37 + 1}}{2}} \right)^{th}}$ term, that is, ${19^{th}}$ term of the sequence is the middle most term.
And each term on either side will include three middle most terms.
Hence, the three middle most terms are ${18^{th}}$ term, ${19^{th}}$ term and the ${20^{th}}$ term.
We know that ${n^{th}}$ term of the sequence is given by $a + \left( {n - 1} \right)d$
Therefore, we have,
${18^{th}}$ term is $a + 17d$, ${19^{th}}$ term is $a + 18d$ and the ${20^{th}}$ term is $a + 19d$
Then, the sum of these terms is given as 225.
Thus,
$a + 17d + a + 18d + a + 19d = 225 \\\ \Rightarrow 3a + 54d = 225 \\\$
On dividing the equation throughout by 3, we get,
$\Rightarrow a + 18d = 75$ (1)
Also, the sum of last three terms is 429
The last three terms will be ${37^{th}}$ term, ${36^{th}}$ term and the ${35^{th}}$ term.
Then, ${37^{th}}$ term is $a + 36d$, ${36^{th}}$ term is $a + 35d$ and the ${35^{th}}$ term is $a + 34d$
Then, we have,
$a + 36d + a + 35d + a + 34d = 429 \\\ \Rightarrow 3a + 105d = 429 \\\$
On dividing the equation throughout by 3, we get,
$\Rightarrow a + 35d = 143$ (2)
Subtract equation (1) and (2) to find the value of $d$
$a + 18d - a - 35d = 75 - 143 \\\ \Rightarrow - 17d = - 68 \\\ \Rightarrow d = 4 \\\$
Substitute the value of $d$ in equation (1) and find the value of $a$
$a + 18\left( 4 \right) = 75 \\\ \Rightarrow a + 72 = 75 \\\ \Rightarrow a = 3 \\\$
Hence, the first term of the A.P. is 3 and the common difference is 4.
Also, the A.P. is given by $a,a + d,a + 2d + ....$
Here, the A.P. is $3,7,11,15,....$
But we have to find the sum of all the 37 terms of A.P.
And we know that the sum of $n$terms is given by ${S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$
We have $n = 37$, $a = 3$ and $d = 4$
On substituting the values, we get,
${S_{37}} = \dfrac{{37}}{2}\left( {2\left( 3 \right) + \left( {37 - 1} \right)4} \right) \\\ \Rightarrow {S_{37}} = \dfrac{{37}}{2}\left( {6 + \left( {36} \right)4} \right) \\\ \Rightarrow {S_{37}} = \dfrac{{37}}{2}\left( {150} \right) \\\ \Rightarrow {S_{37}} = 2,775 \\\$
Hence, the sum of 37 terms of the A.P. whose first term is 3 and common difference is 4 is 2,775.

Note The sum can also be calculated using the formula, ${S_n} = \dfrac{n}{2}\left( {a + {a_n}} \right)$, for using this formula, we will also have to calculate the value of ${a_n}$ using the formula ${a_n} = a + \left( {n - 1} \right)d$. Also, substitute the values correctly to get the correct answer.