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Find the sum of first seventeen terms of an AP whose 4th and 9th terms are - 15 and - 30 respectively
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Your question is....
Find the sum of first seventeen terms of an AP whose 4th and 9th terms are - 15 and - 30 respectively....
Your answer is...
Let the first term be common difference and the number of term in an AP are a, d and n....
So, We know that the nth term of an AP....
☞ Tn = a + ( n - 1 ) d.............(i)
•°• 4th term of an AP.....
☞ T4 = a + ( 4 - 1 ) d
☞ T4 = -15 [ Given]
→ a + 3d = - 15.............(ii)
And, 9th term of AP.....
☞ T9 = a + ( 9 - 1 ) d
☞ T9 = -30 [Given]
→ a + 8d = -30.............(iii)
Now, subtract (i) , (ii) , (iii)
Then we get.....
a + 8d = -30
a + 3d = -15
- - +
--------------------
5d = -15
d = -3
Now put the value of D in (i)
Then we get......
☞ a + 3 ( -3 ) => (-15)
☞ a + 3 ( -3 ) => [ a - 9 = -15 ]
☞ a + 3 ( -3 ) => a = -15 + 9
☞ a + 3 ( -3 ) => a = -6
°•° Sum of first n terms of an AP.....
Sn = n / 2 [ 2a + ( n - 1 ) d ]
Sum of 17 terms of an AP.....
S17 = 17 / 2 [ 2 × ( -6 ) + ( 17 - 1 ) ( -3 ) ]
S17 = 17 / 2 [ -2 + ( 16 ) ( -13 )
S17 = 17 / 2 ( -12 -48 )
S17 = 17 / 2 × ( -60 )
S17 = 17 × ( -30 )
S17 = -510
Hence, The required sum of the first 17 term of an AP is -510....