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Question
Calculate the sum of 35 terms in an AP, whose fourth term is 16 and ninth term is 31.
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Solution
Let a and d be the first term and the common difference, respectively.
Now, a4 = 16 ......[Given]
⇒ a + 3d = 16 .....(i)
Also, a9 = 31 ......[Given]
⇒ a + 8d = 31 .....(ii)
Subtracting the equation (i) from equation (ii), we get
a + 8d = 31
a + 3d = 16
– – –
5d = 15
⇒ d = 3
Substituting the value of d in equation (i), w e get
a + 3 (3) = 16
⇒ a + 9 = 16
⇒ a = 16 – 9 = 7
Now, `"S"_n = n/2 [2a + (n - 1)d]`
∴ `"S"_35 = 35/2 [2(7) + (35 - 1)3]`
= `35/2 [14 + (34)3]`
= `35/2 (14 + 102)`
= `35/2 (116)`
= 35 × 58
= 2030
Hence, the sum of 35 terms of the A.P. is 2030.
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