Question

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

Answer 1

We are dealing with a problem here, where we are to find the arithmetic progression from a given problem. So, we will start with forming our A.P in such a way that whenever we add the terms, it turns out to be in such a way that favors our condition. Hence, we get on with considering the middle term as a and the common difference as d. Then, using the conditions given, we initially get the value of a, and then the other condition gives us the value of d.

Complete step by step solution:
According to the given problem, we are to find an A.P with the given conditions.
Now, we are said that the A.P has 37 terms, so the middle term is $=\dfrac{37+1}{2}th\,term=19th\,term$
Then, we let us consider the 19th term as ‘a’ and the common difference being d.
Thus, the A.P is, a – 18d, a – 17d, …….., a, a + d, a + 2d,….. , a+17d, a+18d
It is said that the sum of the three middle most terms is 225.
So, we get, $\left( a-d \right)+a+\left( a+d \right)=225$
Simplifying,
$\Rightarrow 3a=225$
So, we are getting, $a=75$
Again, it is also said, that the sum of last three terms is, 429,
So, we get, $\left( a+16d \right)+\left( a+17d \right)+\left( a+18d \right)=429$
Simplifying, $3a+51d=429$
Putting the value of a, $3\times 75+51d=429$
Again,
$\Rightarrow 225+51d=429$
$\Rightarrow 51d=429-225=204$
So, we get, $d=\dfrac{204}{51}=4$
Thus, we have our arithmetic progression as,
$75-\left( 18\times 4 \right),75-\left( 17\times 4 \right),75-\left( 16\times 4 \right),......,75-4,75,75+4,75+\left( 2\times 4 \right),....,75+\left( 17\times 4 \right),75+\left( 18\times 4 \right)$ By further simplification, we are getting,
$75-72,75-68,75-64,......,75-4,75,75+4,75+8,....,75+68,75+72$
Hence, the arithmetic progression looks like,
3,7,11,….,71, 75, 79, 83, ….,143, 147

Note: This a typical type of A.P problem where we have to start with considering the middle term as ‘a’. As there are an odd number of elements, we are getting the middle term very easily. If the A.P had an even number of terms, the process would not be the same and we had to consider two middle terms.