
There are 37 terms in an A.P. the sum of the middle three terms is 225 and the sum of the last three terms is 429. Find A.P.
Answer
509.1k+ views
Hint: First find the middle term and then form two equations using the two given conditions. This will help in finding the value of common difference d. then using this value of d find the A.P.
Complete step by step answer:
Given that there are 37 terms.
So n=37.
Middle term = \[\dfrac{{n + 1}}{2} = \dfrac{{37 + 1}}{2} = \dfrac{{38}}{2} = 19\]
In order to find \[{n^{th}}\] term in an A.P. we use formula \[a + \left( {n - 1} \right)d\].
Three middle terms are\[{18^{th}},{19^{th}},{20^{th}}\].
But in an A.P. we can write,
\[{18^{th}} = a + 17d\]
\[{19^{th}} = a + 18d\]
\[{20^{th}} = a + 19d\]
Similarly, we can write for the last three terms
\[
{35^{th}} = a + 34d \\
{36^{th}} = a + 35d \\
{37^{th}} = a + 36d \\
\\
\]
The Sum of the three middle terms is 225.
\[
\Rightarrow a + 17d + a + 18d + a + 19d = 225 \\
\Rightarrow 3a + 54d = 225 \\
\]
Sum of last three terms is 429.
\[
\Rightarrow a + 34d + a + 35d + a + 36d = 429 \\
\Rightarrow 3a + 105d = 429 \\
\]
Now from these two equations find the value of d.
\[
\Rightarrow 3a + 105d - \left( {3a + 54d} \right) = 429 - 225 \\
\Rightarrow 105d - 54d = 204 \\
\Rightarrow 51d = 204 \\
\Rightarrow d = \dfrac{{204}}{{51}} \\
\Rightarrow d = 4 \\
\]
Now putting this value in any of the above equation we can find value of a.
\[
\Rightarrow 3a + 54d = 225 \\
\Rightarrow 3a + 54 \times 4 = 225 \\
\Rightarrow 3a = 225 - 216 \\
\Rightarrow 3a = 9 \\
\Rightarrow a = 3 \\
\]
Since we have the value of the first term and that of common difference also. So let’s find the A.P.
Therefore the A.P. series is \[3,7,11,15,....\]
Note: Finding the middle term is the important step of the problem because the whole problem is dependent on it. Once we get the middle term we can use the conditions given in the problem.
Additional information:An arithmetic progression is a series in which two terms are having a common difference d between them. a, a+d, a+2d, ….is an A.P. Sum of first n terms in an A.P. is given by the formula \[{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\].
Complete step by step answer:
Given that there are 37 terms.
So n=37.
Middle term = \[\dfrac{{n + 1}}{2} = \dfrac{{37 + 1}}{2} = \dfrac{{38}}{2} = 19\]
In order to find \[{n^{th}}\] term in an A.P. we use formula \[a + \left( {n - 1} \right)d\].
Three middle terms are\[{18^{th}},{19^{th}},{20^{th}}\].
But in an A.P. we can write,
\[{18^{th}} = a + 17d\]
\[{19^{th}} = a + 18d\]
\[{20^{th}} = a + 19d\]
Similarly, we can write for the last three terms
\[
{35^{th}} = a + 34d \\
{36^{th}} = a + 35d \\
{37^{th}} = a + 36d \\
\\
\]
The Sum of the three middle terms is 225.
\[
\Rightarrow a + 17d + a + 18d + a + 19d = 225 \\
\Rightarrow 3a + 54d = 225 \\
\]
Sum of last three terms is 429.
\[
\Rightarrow a + 34d + a + 35d + a + 36d = 429 \\
\Rightarrow 3a + 105d = 429 \\
\]
Now from these two equations find the value of d.
\[
\Rightarrow 3a + 105d - \left( {3a + 54d} \right) = 429 - 225 \\
\Rightarrow 105d - 54d = 204 \\
\Rightarrow 51d = 204 \\
\Rightarrow d = \dfrac{{204}}{{51}} \\
\Rightarrow d = 4 \\
\]
Now putting this value in any of the above equation we can find value of a.
\[
\Rightarrow 3a + 54d = 225 \\
\Rightarrow 3a + 54 \times 4 = 225 \\
\Rightarrow 3a = 225 - 216 \\
\Rightarrow 3a = 9 \\
\Rightarrow a = 3 \\
\]
Since we have the value of the first term and that of common difference also. So let’s find the A.P.
Therefore the A.P. series is \[3,7,11,15,....\]
Note: Finding the middle term is the important step of the problem because the whole problem is dependent on it. Once we get the middle term we can use the conditions given in the problem.
Additional information:An arithmetic progression is a series in which two terms are having a common difference d between them. a, a+d, a+2d, ….is an A.P. Sum of first n terms in an A.P. is given by the formula \[{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\].
Recently Updated Pages
Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Trending doubts
Difference Between Prokaryotic Cells and Eukaryotic Cells

1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Net gain of ATP in glycolysis a 6 b 2 c 4 d 8 class 11 biology CBSE

Give two reasons to justify a Water at room temperature class 11 chemistry CBSE
