Question

in an A. P. 10th term is 46 ,sum of 5th and 7th term is 52 . Find the A.P.
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Answer 1

Answer:

Sum of the 5th and 7th term is 52. Now, subtracting equation (1) from equation (2), we get. Substituting the value of d in equation (1),we get. Therefore, the required AP is 1, 6, 11, 16, 21, 26, 31...

Answer 2

Answer:

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Step-by-step explanation:

Given:

The 10th term of an A.P is 46 => t10 = 46

$the \: general \: term \: tp \: = a + (n - 1)d$

where a is first term, d is the common difference.

t10 = a + (10 - 1) d

46 = a +9d ____(1)

t5 + t7 = 52

a + 4d + a + 6d = 52

2a + 10 d = 52

Divide by 2,

a + 5d = 26 ____(2)

Simplify the Equations 1 and 2,

Subtract 1 and 2,,

a + 9d = 46

a + 5d = 26 (-)

_____________

4d = 20

d = 5

Sub the value of d in equation (1),

(1)=> a + 9(5) = 46

a +45 =46

a = 46 -45

a = 1

To find:

$a.p \: = a +( a + d) + (a + 2 d) + (a + 3d) + .....$

A.P = 1 + 6 + 11 + 16 +......

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