Please solve these questions properly...
1) The first and last term of the GP are 3 and 96 respectively. If the common ratio is 2 then find:
i) Number of terms
ii) Sum of 'n' terms
2)The numbers whose sum is 21 are in AP. If 2,2,14 are added to them respectively, the resulting numbers are in GP. Find the numbers.
3) If a,b,c,d are in GP, prove that a+b, b+c, c+d are also in GP.
4) If the third term of a GP is 5/2, and it's eighth term is 5/64, then find the sum of it's first 10 terms.
5) If the 5th and 10th terms of a GP are 32 and 1024 respectively, find the first term and common ratio.
Answers
Answer:
1 )(i) given first term(a)=3
last term(T)=96
common ratio(r)=2
last term in GP is ar^(n-1),n is total number of terms..hence,
96=(3)(2)^(n-1)
32=2^(n-1)
2^5=2^(n-1)
bases are equal..hence powers also equal
5=n-1
n=6 terms
(ii)sum of 'n' terms in GP is given by
S=a(r^n-1)/(r-1)
S=3(2^6-1)/(2-1)
S=3(64-1)/1
S=3(63)
S=189
3,6,12,24,48,96 are the numbers that are in GP
2)It is given that the sum of these numbers is 21.
So, a-d+a+a+d = 21
This gives, a = 7
So, the numbers are 7-d, 7, 7+d.
It is given that if 2, 2, 14 respectively are added to the numbers, then the resulting numbers are in GP.
So, 9-d, 9, 21+d are in GP.
Therefore,
9/9-D =21+d/9
by simplyfying this we get
d^2+12d-108=0
=(d+18)(d-6)=0
d=-18 or 6
Thus, when d = -18, the numbers are 25, 7, -11.
When d = 6, the numbers are 1, 7, 13
4)Ar^2=5/2 ar^7=5/64
Therefore r =1/2 (divide the above two)
Find a by putting r in anyone equation
A=10
S10=a(1-r^n)/1-r
Cause r is smaller than 1
Therefore sum is 20(1-1/2^10)
Sum = 19.98046875
I ONLY KNOW QUESTION 1,2 AND 4. HOPE IT HELPS