The sum of the 1st 13 terms of an Arithmetic sequence is 208 & the sum of the 1st 16 terms is 304.
a What is the 7th term
b) 15th term?
c) Find the sum of the terms from the 14th term to the 29th terms
Answers
Step-by-step explanation:
no
Answer:
(i) 7th term = 16
(ii) 15th term = 30
(iii) Sum of the terms from 14th term to 29th term = 688
Step-by-step explanation:
Given,
Sum of 1st 13 terms = S₁₃ = 208
Sum of 1st 6 terms = S₁₆ = 304
We can write,
S₁₃ = (13/2)[2a+(13-1)d] = 208
2a+12d = 208×(2/13)
a+6d = 208/13
a+6d = 16 ---------(i)
S₁₆ = (16/2)[2a+(16-1)d] = 304
8×[ 2a+15d] = 304
2a+15d = 304/8
2a+15d = 38 ------------(ii)
multiply equation (i) with 2
2[a+6d] = 16×2
2a+12d = 32 -----------------(ii)
Subtract equation (ii) and (iii) then we get
15d-12d = 38-32
3d = 6
d = 2
Now, substitute d = 2 in equation(i) to get 'a' value.
a+6(2) = 16
a = 16-12
a = 4
we have a = 4 and d = 2
The first term a = 4 and the common difference d = 2 for the Arithmetic progression.
(i) The 7th term = a₇ = a+6d = 4+6(2) = 4+12 = 16
So, 7th term = 16
(ii) 15th term = a₁₅ = a+14d = 2+14(2) = 2+28 = 30
So, 15th term = 30
(iii) Sum of the terms from 14th term to 29th term
=[Sum of first 29 terms - Sum of first 14terms] - common difference
= [S₂₉ - S₁₄] - 2
= {(29/2)(2a+(29-1)d)} - {(14/2)(2a+(14-1)d)} - 2
= [(29/2)(2(4)+28(2))] - [7(2(4)+13(2))] - 2
= [(29)(4+28)] - [ 7(8+26)] - 2
= [(29)(32)] - [(7)(34)] - 2
= (928 - 238 ) - 2
= 690 - 2
= 688
Sum of the terms from 14th term to 29th term = 688
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