Question

An arithmetic progression consists of 21 terms. The sum of the three terms in the middle is 63 and the sum of the last three terms is 90. Find the arithmetic progression.

Answer 1

Hint: First determine the number of middle most term by using the formula of middle term of an arithmetic progression (different for both odd and even value of n).Now use nth term formula to write three middle most term and make an equation using the data given in the question , similarly make another equation of last three terms. Now solve these two equations.

Complete step by step solution:
To solve this problem, we need to first enlist the conditions in terms of the constraints given. The first condition is that the sum of the three terms in the middle is 63. The middle terms are given by –
$\dfrac{n+1}{2}-1,\dfrac{n+1}{2},\dfrac{n+1}{2}+1$
Thus, for n = 21, the middle terms are – ${{10}^{th}}$ term, ${{11}^{th}}$ term and ${{12}^{th}}$ term.
Now, using the formula for the ${{n}^{th}}$ term of the arithmetic progression, which is given by –
${{a}_{n}}$= a + (n-1) d
${{a}_{n}}$= ${{n}^{th}}$ term of this series
a = first term of the series
n = number of terms in the series (here there are 21 terms)
d= common difference
Now, we have,
(a + 9d) + (a + 10d) + (a + 11d) = 63
3a + 30d = 63
a + 10d = 21 -- (1)
Further, we also know that sum of the least three terms is 90, thus, we have,
(a+18d) + (a+19d) + (a+20d) = 90
3a + 57d = 90
a + 19d = 30 -- (2)
Now, solving (1) and (2), we get,
d = 1, a = 11
Thus, the arithmetic progression is – 11, 12, 13, …, 31.

Note: To find the arithmetic progression, we need three independent conditions to find the entire series. In this case, we know the total number of terms are 21 and in addition we also have two other conditions, thus we were able to solve the problem. We can thus use the known formulas of arithmetic progression to solve the problem.