CA Foundation Exam > CA Foundation Questions > t12 of the series –128, 64, –32, ...
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t12 of the series –128, 64, –32, ….is
- a)– 1/16
- b)16
- c)1/16
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
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t12 of the series –128, 64, –32, ….isa)– 1/16b...
Hi there ... I hope you find it easy .
You see these numbers have something to do with 2^n .
128=2^7 , 64= 2^6 and so on.
This is a Geometric Progression. So the formula is A(n)=A.r^(n-1).
Here A(n) = A(12) A=128 and r=1/2 .
(You can find r by taking ratio of 2nd term to 1st and confirming by doing the same for 3rd to 2nd term . Which here gives 1/2)
Thus putting values in A(n)=A.r^n-1 gives A(12)=128.(1/2)^12-1which ends us up giving :-
1/16.
Most Upvoted Answer
t12 of the series –128, 64, –32, ….isa)– 1/16b...
Geometric Progression (GP)
The given series 128, 64, 32, ... is a Geometric Progression (GP) because each term is obtained by multiplying the previous term by a constant factor of 1/2.
Formula for nth term of a GP
The formula for finding the nth term of a GP is:
an = a1 * r^(n-1)
where
an = nth term
a1 = first term
r = common ratio
n = number of terms
Finding the common ratio (r)
In this GP, the first term (a1) is 128 and the second term (a2) is 64. We can use these two values to find the common ratio (r) as follows:
r = a2 / a1
r = 64 / 128
r = 1/2
Using the formula to find t12
We can now use the formula for finding the nth term of a GP to find t12 (the 12th term) as follows:
t12 = a1 * r^(12-1)
t12 = 128 * (1/2)^11
t12 = 128 * 1/2048
t12 = 1/16
Therefore, the 12th term of the series 128, 64, 32, ... is 1/16. Hence, the correct answer is option C.
The given series 128, 64, 32, ... is a Geometric Progression (GP) because each term is obtained by multiplying the previous term by a constant factor of 1/2.
Formula for nth term of a GP
The formula for finding the nth term of a GP is:
an = a1 * r^(n-1)
where
an = nth term
a1 = first term
r = common ratio
n = number of terms
Finding the common ratio (r)
In this GP, the first term (a1) is 128 and the second term (a2) is 64. We can use these two values to find the common ratio (r) as follows:
r = a2 / a1
r = 64 / 128
r = 1/2
Using the formula to find t12
We can now use the formula for finding the nth term of a GP to find t12 (the 12th term) as follows:
t12 = a1 * r^(12-1)
t12 = 128 * (1/2)^11
t12 = 128 * 1/2048
t12 = 1/16
Therefore, the 12th term of the series 128, 64, 32, ... is 1/16. Hence, the correct answer is option C.
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t12 of the series –128, 64, –32, ….isa)– 1/16b)16c)1/16d)none of theseCorrect answer is option 'C'. Can you explain this answer?
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