Question

In an arithmetic progression 3rd term is 10 and 7th term is 22, find its 10" term​

Answer 1

Answer:

a3 = 10

=> a + (n-1)d= 10

=> a + (3-1)d= 10

=> a + 2d = 10 ...( 1 )

similarly,

a7 = 22

=> a+ 6d = 22 ...( 2 )

subtracting ( 2 ) from ( 1 )

4d = 12

d = 4

putting value of d in (1)

a + 2(4) = 10

=> a = 2

now,

a10 = a+ (n- 1) d

=> 2 + (10 - 1) 4

=> 2+ 9 (4)

=> 2 + 36

=> 38

it's 10th term is 38

Answer 2

Given

$\sf{ \to a_3 = 10 }\\ \sf{ \to a_7 = 22}$

To find

$\sf{ \to a_{10}=?}$

Concept used

Here the concept of Arithmetic progression is used. First we will expand the given terms in form of (a+n×d) then solve them to get the value of d(common difference). Then after subsitute the value of d in any of the equation to get the value of a. Súbsitúte the value of a and d in a+9d.

Solution

$\sf{ \to a_3= a+2d = 10}\\ \\ \sf{ \to a_7= a+ 6d= 22}$

Subtract both the equatîons

$\sf{a+2d= 10}\\ \sf{(-)a+6d= 22} \\ \sf{--------}\\\sf{ \implies-4d= -12}\\ \sf{ \implies d =\frac{12}{4}}\\ \sf \implies\red{d= 3}$

Now súbsitúte the value of d in any of the equation.

$\sf{ \implies a+2d=10}\\ \\ \sf{a+ 2×3 =10 } \\ \\\sf{a+ 6=10 }\\ \\ \sf \green{a= 4}$

$\mathtt{ \bull Now\:we\: have \: to \: find\:a_{10}\: term.}$

$\sf{ \to a_{10} = a+ 9d}$

súbsitúte the value of a and d

$\sf{ \implies a_{10 }= 4 + 9× 3} \\ \\ \sf{ \implies a_{10} = 4+27}\\ \\ \sf{\implies \red{ a_{10} = 31}}$

Therefore the 10th term is 31.

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