the 17 term of AP is 5
more than twice its 9 term. If the 11 term of AP 43 find the no. of terms and the sum of the number
Answers
Answered by
0
Step-by-step explanation:
we know that the n th
term of the arithmetic progression is given by a+(n−1)d
Given that the 17 th
term is 5 more than twice the 8 th
term.
Therefore, 5+17 th term=2(8 th term)
⟹a+(17−1)d=2(a+(8−1)d)+5
⟹a+16d−5=2a+14d
⟹a−2d=−5 -------(1)
Given that the 11th term is 43
Therefore, a+(11−1)d=43
⟹a+10d=43 ------(2)
subtracting (1) from (2) we get
(a+10d)−(a−2d)=43−(−5)
⟹12d=48
⟹d=4
⟹a=3
Therefore, the n th term is 3+(n−1)4=4n−1
Similar questions