Question

The first term of an A.P. is 14 and the sums of the first five terms and the first ten terms are equal in magnitude but opposite in sign. Find the third term of the progression.

Answer 1

Hint: The sum of n terms of A.P. means the sum of first n terms of the arithmetic progression. Using the formula to find out the sum of the first n terms of an A.P., find out the sum of the first five terms of given A.P. and equate it with the negative of the sum of the first ten terms of the same A.P., this way you will obtain the common difference of the A.P. and can easily find out any term of the A.P.

Complete step-by-step answer:
The first term of A.P. is 14, $ a = 14 $
Sum of n terms of an A.P. is given as $ {S_n} = \dfrac{n}{2}[2a + (n - 1)d] $
The sum of the first five terms is $ {S_5} = \dfrac{5}{2}[2(14) + (5 - 1)d] = \dfrac{5}{2}(28 + 4d) = 5(14 + 2d) $
The sum of the first ten terms is $ {S_{10}} = \dfrac{{10}}{2}[2(14) + (10 - 1)d] = 5(28 + 9d) $
Now, we are given that sum of the first five terms is equal in magnitude with the sum of the first ten terms but opposite in sign, so –
 $
\Rightarrow 5(14 + 2d) = - 5(28 + 9d) \\
\Rightarrow 14 + 2d = - 28 - 9d \\
\Rightarrow 11d = - 42 \\
\Rightarrow d = \dfrac{{ - 42}}{{11}} \;
  $
 $ {n^{th}} $ term of an A.P. is given as $ {a_n} = a + (n - 1)d $
So the third term of A.P. is $ {a_3} = 14 + (3 - 1)(\dfrac{{ - 42}}{{11}}) = 14 + (\dfrac{{ - 84}}{{11}}) = 14 - \dfrac{{84}}{{11}} = \dfrac{{70}}{{11}} $
Hence, the third term of the progression is $ \dfrac{{70}}{{11}} $ .
So, the correct answer is “ $ \dfrac{{70}}{{11}} $ ”.

Note: An arithmetic progression is a progression or sequence of numbers such that the difference between any two consecutive numbers is constant.
For example, the sequence 5, 9, 13, 17…. is an arithmetic progression or arithmetic sequence whose common difference can be obtained by subtracting one term from its next term, the common difference in this example is 4.