Question

The sum of first 6 terms of an arithmetic sequence is 99. Its 6 th term is 39. (1). Find the sum of 3 rd and 4 th term? (2). which is the 3 rd term? (3). write the sequence?

Answer 1

Step-by-step explanation:

answer

1. is 33

2. is 12

3. is -6 , 3 , 12 , 21 , 30 , 39 , 48 ..........

Answer 2

Given - The sum of the first 6 terms of an arithmetic sequence is 99. Its 6 th term is 39. (1).

Find the sum of 3 rd and 4 th term? (2). which is the 3 rd term? (3) write the sequence?

An arithmetic progression or arithmetic sequence (AP) is a sequence of numbers such that the difference between the consecutive terms is constant. For instance, the sequence 5, 7, 9, 11, 13, 15, . . . is an arithmetic progression with a common difference of 2.

If the initial term of an arithmetic progression is a and the common difference of successive members is d, then the n-th term of the sequence $a_n$ is given by:

$a_n=a+(n-1)d$

If there are m terms in the AP, then $a_{m}$ represents the last term which is given by:

${\displaystyle a_{m}=a+(m-1)d}$

A finite portion of an arithmetic progression is called a finite arithmetic progression and sometimes just called an arithmetic progression. The sum of a finite arithmetic progression is called an arithmetic series.

$S_6=99$

$a_6=39$$a_6=39$

From $a_6=39$

$a+5d=39$.....(i)

$\Rightarrow \frac{6}{2}[2a+(6-1)d]=99$

$\Rightarrow 3[2a+5d]=99$  $[\therefore S_n=\frac{n}{2}[2a+(n-1)d]$

$\Rightarrow 2a+5d=\frac{99}{3}=33$

$\Rightarrow 2a+5d=33$......(ii)

Subtracting (i) from (ii)

$2a+5d=33\\a+5d=39$

$a=-6$

Put $a=-6$ in (i)

$-6+5d=39$

$5d=39+6$

$5d=45$

$d=9$

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