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Question

Question 27
Find the sum of first 17 terms of an AP whose 4th and 9th terms are – 15 and –30, respectively.

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Solution

Let the first term, common difference and the number of terms in the AP are a,d and n, respectively.
We know that, the nth term of an AP, Tn = a + (n - 1)d . . . (i)
4th term of the AP, T4 = a + (4 - 1)d = - 15 [given]
a + 3d = -15 . . . . . (ii)
9th term of the AP, T9 = a + (9 - 1)d = - 30
[given]
a + 8d = - 30 ….(iii)
Now, subtract, eq.(ii) from eq.(iii), we get;
a+8d=30a+3d=15 + –––––––––––5d=15d=3

Put the value of d in eq.(ii), we get;
a + 3 (-3) = - 15 a - 9 = - 15
a = - 15 + 9 a = - 6
\Sum of first n terms of an AP, Sn=n2[2a+(n1)d]
Sum of first 17 terms of the AP, S17=172[2×(6)+(171)(3)]
=172[12+(16)(3)]
=172(1248)=172×(60)
=17×(30)=510

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