The number of terms of an AP is even; the sum of the odd terms is 3075 sum of the even terms is 3200, and the last term exceeds the second term by 240. Find number of terms and series...
Answers
Given that number of terms of an AP is even.
Let the number of terms be a,a + d,a + 2d,..... 2n terms.
(i)
Given that sum of odd terms is 3075.
= > a + (a + 2d) + (a + 4d)....+n = 3075
= > 3075 = (n/2){2a + (n - 1) * 2d} ------ (1)
(ii)
Given that Sum of even terms is 3200.
= > (a + d) + (a + 3d) + .... n terms = 3200
= > 3200 = (n/2){2(a + d) + (n - 1) * 2d} ------ (2)
(iii)
Given that last term exceeds the second term by 240
= > {a + (2n - 1) * d} - (a + d) = 240
= > a + (2n - 1) * d - a - d
= > {(2n - 1) * d} - d = 240 ----- (3)
On solving (1) & (2), we get
= > 125 = n/2[2d]
= > d = 125/n ----- (4)
Substitute d = 125/n in (3), we get
= > {(2n - 1) * 125/n} - 125/n = 240
= > 125(2n - 1) - 125 = 240n
= > 250n - 250 = 240n
= > 10n = 250
= > n = 25.
Therefore, the number of terms = 2 * 25 = 50.
Substitute n = 50 in (4), we get
= > d = 125/50
= > d = 125/50
= > d = 2.5
Substitute d = 2.5, n = 25 in (1), we get
= > 3075 = (25/2)[2a + (25 - 1) * 2 * 2.5]
= > 3075 = (25/2)[2a + 120)
= > 6150 = 25[2a + 120]
= > 2a + 120 = 246
= > 2a = 126
= > a = 63.
Therefore, number of terms = 50.
Number of series = 63, 65.5 ,68...
Hope this helps!