Question

The ratio of the fifth term to the twelfth term of a sequence in an arithmetic progression is 6/13. If each term of this sequence is positive, and the product of the first term and the third term is 32, find the sum of the first 100 terms of this sequence.

Answer 1

Step-by-step explanation:

11th

Maths

Sequences and Series

Arithmetic Progression

(a) The fifth term of an ar...

MATHS

(a) The fifth term of an arithmetic sequence is 40 and tenth term is 20. What is the fifteenth term?

(b) How many terms of this sequence make the sum zero?

MEDIUM

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VIDEO EXPLANATION

ANSWER

(a) x

5

=40,x

10

=20

x

10

−x

5

=5d

5d=20−40=−20

⇒d=−4

Now, x

15

=x

10

+5d=20−20=0

(b) First term, f=x

5

−4d=40−4(−4)=56

S

n

=

2

n[2f+(n−1)d]

⇒0=

2

n[2×56+(n−1)(−4)]

⇒n(112−4n+4)=0

⇒n(116−4n)=0

⇒=0,n=29

Thus, 29 terms of this sequence make the sum zero.

Answer 2

Answer:

The sum of the first $100$ terms of this sequence is $20,000$

Step-by-step explanation:

Given:

The ratio of the fifth term to the twelfth term of a sequence is an arithmetic progression is $\frac{6}{13}$

the product of the first term and the third term is $32$

We need to find the sum of the first $100$ terms of this sequence.

Let's write the given statements in mathematical form as

$\frac{a+4d}{a+11d}=\frac{6}{13}$--(1)

And

$a(a+2d)=32$---(2)

Solving first equation we get as

$13(a+4d)=6(a+11d)\\\\13a+52d=6a+66d\\\\7a=14d\\\\a=2d$

Substituting the condition in second equation we get

$a(a+a)=32\\\\2a^2=32\\\\a^2=16\\\\a=4$as given $a>0$

The common difference will be

$a=2d\\\\2d=4\\\\d=2$

Sum of n terms in an A.P is

$\frac{n}{2}[2a+(n-1)d]$

By substituting the values we get as

$S_{100}=\frac{100}{2}[2(2)+(99)(4)]\\\\S_{100}=50[400]\\\\S_{100}=20,000$