Question -
Answer -
(i) 2, 6, 18, … to 7 terms
We know that, sum ofGP for n terms = a(rn – 1)/(r – 1)
Given:
a = 2, r = t2/t1 =6/2 = 3, n = 7
Now let us substitutethe values in
a(rn –1)/(r – 1) = 2 (37 – 1)/(3-1)
= 2 (37 –1)/2
= 37 –1
= 2187 – 1
= 2186
(ii) 1, 3, 9, 27, … to 8terms
We know that, sum ofGP for n terms = a(rn – 1)/(r – 1)
Given:
a = 1, r = t2/t1 =3/1 = 3, n = 8
Now let us substitutethe values in
a(rn –1)/(r – 1) = 1 (38 – 1)/(3-1)
= (38 –1)/2
= (6561 – 1)/2
= 6560/2
= 3280
(iii) 1, -1/2, ¼, -1/8, …
We know that, sum ofGP for infinity = a/(1 – r)
Given:
a = 1, r = t2/t1 =(-1/2)/1 = -1/2
Now let us substitutethe values in
a/(1 – r) = 1/(1 –(-1/2))
= 1/(1 + 1/2)
= 1/((2+1)/2)
= 1/(3/2)
= 2/3
(iv) (a2 –b2), (a – b), (a-b)/(a+b), … to n terms
We know that, sum ofGP for n terms = a(rn – 1)/(r – 1)
Given:
a = (a2 –b2), r = t2/t1 = (a-b)/(a2 –b2) = (a-b)/(a-b) (a+b) = 1/(a+b), n = n
Now let us substitutethe values in
a(rn –1)/(r – 1) =
(v) 4, 2, 1, ½ … to 10terms
We know that, sum ofGP for n terms = a(rn – 1)/(r – 1)
Given:
a = 4, r = t2/t1 =2/4 = 1/2, n = 10
Now let us substitutethe values in
a(rn –1)/(r – 1) = 4 ((1/2)10 – 1)/((1/2)-1)
= 4 ((1/2)10 –1)/((1-2)/2)
= 4 ((1/2)10 –1)/(-1/2)
= 4 ((1/2)10 –1) × -2/1
= -8 [1/1024 -1]
= -8 [1 – 1024]/1024
= -8 [-1023]/1024
= 1023/128