CA Foundation Question > The sum of a certain number of terms of an AP...

The sum of a certain number of terms of an AP series –8, –6, –4, …… is 52. The number of terms is

- a)12
- b)13
- c)11
- d)none of these

Correct answer is option 'B'. Can you explain this answer?

a = -8,

d =2,

Sn= 52,

we know that,

Sn=n/2[2a+(n-1)d]

52=n/2[-16+2n-2]

n(n-9)=52

n^2-9n-52=0

(n-13)(n+4)=0

n=13 since n=-4 is impossible.

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The sum of a certain number of terms of an AP series –8, –6, –4, …… is 52. The number of terms isa)12b)13c)11d)none of theseCorrect answer is option 'B'. Can you explain this answer?

Formula is n/2(2a+(n-1)d)

so n/2(2(-8)+(n-1)2)=n/2(-16+2n-2)

=n/2(-18+2n)=52

-18n+2n^2=104

n^2-9n-52=0

n^2-13n+4n-52=0

n(n-13)+4(n-13)=0

(n-13)(n+4)=0

n=13. n=-4

Here we consider only positive values so n=13

Hence it is proved

so n/2(2(-8)+(n-1)2)=n/2(-16+2n-2)

=n/2(-18+2n)=52

-18n+2n^2=104

n^2-9n-52=0

n^2-13n+4n-52=0

n(n-13)+4(n-13)=0

(n-13)(n+4)=0

n=13. n=-4

Here we consider only positive values so n=13

Hence it is proved

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The sum of a certain number of terms of an AP series –8, –6, –4, …… is 52. The number of terms isa)12b)13c)11d)none of theseCorrect answer is option 'B'. Can you explain this answer? for CA Foundation 2023 is part of CA Foundation preparation. The Question and answers have been prepared according to the CA Foundation exam syllabus. Information about The sum of a certain number of terms of an AP series –8, –6, –4, …… is 52. The number of terms isa)12b)13c)11d)none of theseCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for CA Foundation 2023 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The sum of a certain number of terms of an AP series –8, –6, –4, …… is 52. The number of terms isa)12b)13c)11d)none of theseCorrect answer is option 'B'. Can you explain this answer?.

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a = -8,d =2,Sn= 52,we know that,Sn=n/2[2a+(n-1)d]52=n/2[-16+2n-2]n(n-9)=52n^2-9n-52=0(n-13)(n+4)=0n=13 since n=-4 is impossible.