Question -
Answer -
(i) Given:
AP: 7, 10, 13,…
Here, a1 =a = 7, a2 = 10
Common difference, d =a2 – a1 = 10 – 7 = 3
We know, an =a + (n – 1) d [where a is first term or a1 and d is commondifference and n is any natural number]
an = 7+ (n – 1)3
= 7 + 3n – 3
= 3n + 4
To find total terms ofthe A.P., put an = 43 as 43 is last term of A.P.
3n + 4 = 43
3n = 43 – 4
3n = 39
n = 39/3
= 13
Hence, total 13 termsexists in the given A.P.
(ii) Given:
AP: -1, -5/6, -2/3,-1/2, …
Here, a1 =a = -1, a2 = -5/6
Common difference, d =a2 – a1
= -5/6 – (-1)
= -5/6 + 1
= (-5+6)/6
= 1/6
We know, an =a + (n – 1) d [where a is first term or a1 and d is commondifference and n is any natural number]
an =-1 + (n – 1) 1/6
= -1 + 1/6n – 1/6
= (-6-1)/6 + 1/6n
= -7/6 + 1/6n
To find total terms ofthe AP,
Put an =10/3 [Since, 10/3 is the last term of AP]
an =-7/6 + 1/6n = 10/3
1/6n = 10/3 + 7/6
1/6n = (20+7)/6
1/6n = 27/6
n = 27
Hence, total 27 termsexists in the given A.P.