Question -
Answer -
(i)┬а2, 6, 18, тАж to 7 terms
We know that, sum ofGP for n terms = a(rn┬атАУ 1)/(r тАУ 1)
Given:
a = 2, r = t2/t1┬а=6/2 = 3, n = 7
Now let us substitutethe values in
a(rn┬атАУ1)/(r тАУ 1) = 2 (37┬атАУ 1)/(3-1)
= 2 (37┬атАУ1)/2
= 37┬атАУ1
= 2187 тАУ 1
= 2186
(ii)┬а1, 3, 9, 27, тАж to 8terms
We know that, sum ofGP for n terms = a(rn┬атАУ 1)/(r тАУ 1)
Given:
a = 1, r = t2/t1┬а=3/1 = 3, n = 8
Now let us substitutethe values in
a(rn┬атАУ1)/(r тАУ 1) = 1 (38┬атАУ 1)/(3-1)
= (38┬атАУ1)/2
= (6561 тАУ 1)/2
= 6560/2
= 3280
(iii)┬а1, -1/2, ┬╝, -1/8, тАж
We know that, sum ofGP for infinity = a/(1 тАУ r)
Given:
a = 1, r = t2/t1┬а=(-1/2)/1 = -1/2
Now let us substitutethe values in
a/(1 тАУ r) = 1/(1 тАУ(-1/2))
= 1/(1 + 1/2)
= 1/((2+1)/2)
= 1/(3/2)
= 2/3
(iv)┬а(a2┬атАУb2), (a тАУ b), (a-b)/(a+b), тАж to n terms
We know that, sum ofGP for n terms = a(rn┬атАУ 1)/(r тАУ 1)
Given:
a = (a2┬атАУb2), r = t2/t1┬а= (a-b)/(a2┬атАУb2) = (a-b)/(a-b) (a+b) = 1/(a+b), n = n
Now let us substitutethe values in
a(rn┬атАУ1)/(r тАУ 1) =
(v)┬а4, 2, 1, ┬╜ тАж to 10terms
We know that, sum ofGP for n terms = a(rn┬атАУ 1)/(r тАУ 1)
Given:
a = 4, r = t2/t1┬а=2/4 = 1/2, n = 10
Now let us substitutethe values in
a(rn┬атАУ1)/(r тАУ 1) = 4 ((1/2)10┬атАУ 1)/((1/2)-1)
= 4 ((1/2)10┬атАУ1)/((1-2)/2)
= 4 ((1/2)10┬атАУ1)/(-1/2)
= 4 ((1/2)10┬атАУ1) ├Ч -2/1
= -8 [1/1024 -1]
= -8 [1 тАУ 1024]/1024
= -8 [-1023]/1024
= 1023/128