If the sum to infinity of the series 3+(3+d) (1)/(4)+(3+2 d) (1)/(4^2)+…is (44)/(9), then d= (A)

If the sum to infinity of the series 3+(3+d) (1)/(4)+(3+2 d)...

Key Concepts

Infinite Geometric Series

An infinite geometric series consists of terms that have a constant ratio between successive terms. If the absolute value of this ratio is less than one, the series converges, and its sum is calculated using the formula S = a / (1 - r), where 'a' is the first term and 'r' is the common ratio.

Arithmetic-Geometric Series

An arithmetic-geometric series involves terms that are the product of an arithmetic sequence and a geometric sequence. In such series, one part typically has a linear progression while the other part scales geometrically. The sum can often be determined by separating the series into parts that can be individually summed using formulas for geometric series and related summations.

Series Decomposition

Series decomposition is the process of breaking down a complicated series into simpler components that can be individually handled using standard summation formulas. This method is particularly useful when a series combines elements of both arithmetic and geometric progressions, allowing each piece to be summed before combining the results to form the overall sum.

Transcript

00:01 Hello students, today i will discuss this problem in which we have some series whose sum is given as 44 by 9 and we need to find the value of d.

00:10 So if see closely this series, this is 3 into 1 plus 3 plus d into 1 plus 3 plus d into 1 by 4 plus 3 plus 2d into 1 upon 4 square.

00:31 And does does, dash, dash, dash, ds, infinity.

00:34 Let us consider it as s.

00:36 If we see closely, this is ap.

00:43 And its first term is consist of nap and second term would be a gp.

00:50 So we can easily say that this is n -a -g -p.

00:56 And we have a process to find its sum easily by multiplying common ratio of its gp.

01:06 So if we see here common ratio of gp is 1 by 4, so we multiply on both side to get the sum of this series.

01:18 So it would be as upon 4 and we write term 1 ahead so that calculation would be easy.

01:28 So if we multiply 3 into 1 by 1 by 4, it would be 3 into 1 by 4 and this term would be multiplied by 1 by 4 it would be 3 plus d into 1 upon 4 square and dash dash dash if we subtract this so we will have minus sign minus sign so this will be would be 3 s by 4 equal to this would be 3 plus 3 plus d into 1 by 4 minus 3 into 1 by 4 would be only d by 4 and if we talk about this term so from here it would be d upon 4 square and if we think about next term so it would be d upon 4 cube and till till till till infinity we can easily get it into this.

02:34 Now for next step, what we have to do is this would be 3 s upon 4 equal to 3 plus...

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