Question

In an arithmatic progression the sum of 3rd term and 6th term is 28 and the sum of 4 and 8 term is 34 find the A P

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Answer 1

GIVEN,

sum of third and fifth term is 24

sum of fourth and sixth term is 34

an=a+(n-1)d

third term can be written as

a+2d+a+5d=28=2a+7d=28 EQ 1

similarly

a+3d+a+7d=34

=2a+10d=34 EQ 2

solve 1 and 2

2a+7d=28

2a+10d=34

substarct the eq

-3d=-6=d=-6/-3=2

substitute in EQ 1

2a+7(2)=28

2a+14=28

2a=28-14

2a=14=a=14/2=7

a=7,d=2

a,a+d,a+3d you can find

Answer 2

Question :-

In an arithmetic progression, the sum of 3rd term and 6th term is 28 and the sum of 4th and 8 term is 34. Find the A. P. series.

$\large\underline{\sf{Solution-}}$

Let assume that first term of an AP series is a and common difference of an AP is d.

According to statement, it is given that sum of $3^{rd}$term and $6^{th}$ term is 28.

$\rm \: a_3 + a_6 = 28$

$\rm \: a + (3 - 1)d + a + (6 - 1)d = 28$

$\rm \: a + 2d + a + 5d = 28$

$\rm\implies \:\rm \: 2a + 7d = 28 - - - (1)$

According to statement again, the sum of $4^{th}$ term and $8^{th}$ term is 34.

$\rm \: a_4 + a_8 = 34$

$\rm \: a + (4 - 1)d + a + (8 - 1)d = 34$

$\rm \: a + 3d + a + 7d = 34$

$\rm\implies \:\rm \: 2a + 10d = 34 - - - (2)$

On Subtracting equation (1) from equation (2), we get

$\rm \: 3d = 6$

$\rm\implies \:d \: = \: 2$

On substituting the value of d, in equation (1), we get

$\rm \: 2a + 7 \times 2 = 28$

$\rm \: 2a + 14 = 28$

$\rm \: 2a = 28 - 14$

$\rm \: 2a =14$

$\bf\implies \:a \: = \: 7$

Hence,

$\rm\:Required\:AP\:is\:7, 9, 11, 13, 15,\cdots\cdots$

$\rule{190pt}{2pt}$

Formula Used :-

↝ nᵗʰ term of an arithmetic sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{a_n\:=\:a\:+\:(n\:-\:1)\:d}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

• aₙ is the nᵗʰ term.

• a is the first term of the sequence.

• n is the no. of terms.

• d is the common difference.