Question

Consider the arithmetic sequence $6, 10, 14, ....$
(a) What is the sum of first n consequence terms of the above sequence?
(b) How many consecutive terms from the beginning should be added to get the sum 240?
(c) Is the sum of first-few consecutive terms becomes 250? Why?

Answer 1

Hint: In this question, we are given an arithmetic progression. The difference between any two terms is known as the common difference. The difference between any two terms of the given A.P. is 4, so all the parts of this question can be easily solved by using the formula for finding the sum of first n terms of an A.P.

Complete step-by-step solution:
a) The first term of the given A.P. is 6 and its common difference is 4.
We know that the sum of n terms of an A.P. is given as –
$
  {s_n} = \dfrac{n}{2}(2a + (n - 1)d) \\
   \Rightarrow {s_n} = \dfrac{n}{2}[2(6) + (n - 1)4] \\
   \Rightarrow {s_n} = \dfrac{n}{2}(12 + 4n - 4) \\
   \Rightarrow {s_n} = \dfrac{n}{2}(8 + 4n) \\
   \Rightarrow {s_n} = 2{n^2} + 4n \\
 $
Hence, the sum of the first n consequence terms of the given sequence is $2{n^2} + 4n$ .

b) Let the sum of the first m terms be equal to 240, we get –
$
  2{n^2} + 4n = 240 \\
   \Rightarrow {n^2} + 2n = 120 \\
   \Rightarrow {n^2} + 2n - 120 = 0 \\
 $
The obtained equation can be solved by factorization as follows –
$
  {n^2} + 12n - 10n - 120 = 0 \\
   \Rightarrow n(n + 12) - 10(n + 12) = 0 \\
   \Rightarrow (n - 10)(n + 12) = 0 \\
   \Rightarrow n - 10 = 0,\,n + 12 = 0 \\
   \Rightarrow n = 10,\,n = - 12 \\
 $
As the number of terms cannot be negative so -12 is rejected.
Hence, 10 consecutive terms from the beginning should be added to get the sum 240.

c) In this part, we have to find if the sum of the first few consecutive terms is 250 or not. So,
$
  2{n^2} + 4n = 250 \\
   \Rightarrow 2{n^2} + 4n - 250 = 0 \\
   \Rightarrow {n^2} + 2n - 125 = 0 \\
 $
The obtained equation cannot be factorized so we solve it by using the quadratic formula –
$
  n = \dfrac{{ - 2 \pm \sqrt {{{(2)}^2} - 4(1)( - 125)} }}{{2(1)}} \\
   \Rightarrow n = \dfrac{{ - 2 \pm \sqrt {1000} }}{2} \\
   \Rightarrow n = - 1 \pm 5\sqrt {10} \\
 $
The number obtained is not a natural number.
Hence, the sum of the first-few consecutive terms cannot become 250.

Note: An A.P. is defined as a series or sequence of numbers in which the difference between any two consecutive terms is equal. While solving the second and the third part, we got a quadratic equation as the degree of the equation so we solve them by factorization or by applying the quadratic formula.