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Arithmetic Progression Example

Ashish K Sharma
08/06/2022 0 0

                                            Arithmetic Progression Example.

Q:If  the sum of the first 7 terms of an AP is 49 and that of the first 17 terms is 289,find the sum of its first n terms.

 

                                                                  Solution:

 We know that the Sum of the first n terms of an AP is given by:

                                   Sn = (n / 2) * ((2 * a + (n-1) * d)).

 

Therefore, as per the first requirement,

                                          49 = (7 / 2) * ((2 * a + (7-1) * d)).

  • 98 = 14a + 42d
  • 7 = a + 3 d                                    (1)

 

Similarly, as per the second requirement,

                                    289 = (17 / 2) * ((2 * a + (17-1) * d)).

  • 578 = 34a + 272d
  • 17 = a + 8d (2)

 

Now, from (1), we have , a = 7 – 3d. Substituting this value of a in (2), we have

                          17 = 7-3d + 8d

                         Whence, 10 = 5d,

                           Thus, d = 2.

And, then a is 7 – (3 * 2) = 7 – 6 = 1

Thus, we have, a = 1 and d = 2. Now, the sum of the first n terms is

 

                  Sn = (n / 2) * ((2 * a + (n-1) * d)).

  • Sn = (n / 2) * ((2 * 1 + (n-1) * 2)). ---- Substituting the values of a and d obtained earlier
  • Sn = (n / 2) * ((2 + 2n -2))
  • Sn = (n / 2) * (2n)
  • Sn = n^2

Thus,

                    The sum of its first n terms = n^2.

 

 

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