5th term of an arithmetic sequence is 48,16th term is 169.Find the sum of first 21 terms.
Answers
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Answer:
The sum of the first 21 terms of the AP is 2394.
Step-by-step-explanation:
We have given that,
For an arithmetic progression,
t₅ = 48
t₁₆ = 169
We have to find the sum of first 21 terms of the AP.
Now, we know that,
tₙ = a + ( n - 1 ) * d - - - [ Formula ]
⇒ t₅ = a + ( 5 - 1 ) * d
⇒ 48 = a + 4d
⇒ a + 4d = 48
⇒ a = 48 - 4d
⇒ a = - 4d + 48 - - - ( 1 )
Also,
t₁₆ = a + ( 16 - 1 ) * d
⇒ 169 = a + 15d
⇒ a + 15d = 169
⇒ ( - 4d + 48 ) + 15d = 169 - - - [ From ( 1 ) ]
⇒ - 4d + 48 + 15d = 169
⇒ - 4d + 15d + 48 = 169
⇒ 11d + 48 = 169
⇒ 11d = 169 - 48
⇒ 11d = 121
⇒ d = 121 ÷ 11
⇒ d = 11
By substituting d = 11 in equation ( 1 ), we get,
a = - 4d + 48 - - - ( 1 )
⇒ a = - 4 * 11 + 48
⇒ a = - 44 + 48
⇒ a = 4
Now, we know that,
Sₙ = ( n / 2 ) [ 2a + ( n - 1 ) * d ] - - - [ Formula ]
⇒ S₂₁ = ( 21 / 2 ) [ 2 * 4 + ( 21 - 1 ) * 11 ]
⇒ S₂₁ = ( 21 / 2 ) ( 8 + 20 * 11 )
⇒ S₂₁ = ( 21 / 2 ) ( 8 + 220 )
⇒ S₂₁ = ( 21 / 2 ) * 228
⇒ S₂₁ = 21 * 228 ÷ 2
⇒ S₂₁ = 21 * 114
⇒ S₂₁ = 2394
∴ The sum of the first 21 terms of the AP is 2394.