Question

the 10th and 18th terms of an AP are 41 and 73 respectively .find the 26th term?

Answer 1

Hey Friends,
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Let the first term be a & difference be d

Then,
$t_{10} = a + 9d$
$41 = a + 9d$
Again,
$t_{18} = a + 17d$
$73 = a + 17d$
Substract eqn (i) from eqn(ii)
73-41=17d-9d
32=8d
d=4

Put the value of d in eqn (i)
41=a+9×4
41=a+36
a=5

So,
$t_{26} = a + 25d$
$t_{26} = 5 + 25×4$
$t_{26} = 105$

Answer 2

105

Given:

10th term of the AP = 41

18th term of the AP = 73

We can write the 10th term of the AP as:

a + 9d = 41 -------------Equation (1)

a + 17d = 73-----------Equation(2)

To Find:

The 26th term of the AP.

Calculating:

Solving both the linear equations by elimination method:

a + 9d = 41 -------------Equation (1)

a + 17d = 73-----------Equation(2)

   - 8 d = - 32

   d = - 32 / - 8

   d = 4

Therefore, the common difference of the AP is 4.

Now substituting the value of d in equation 1 we get:

a + 9 x 4 = 41

a + 36 = 41

a = 41 - 36

a = 5

Therefore, the first term of the AP is 5.

We can write the 26 th term of the AP as:

a + 25d

Substituting all the values known to us in this we get:

= 5 + 25 x 4

= 5 + 100

= 105

Therefore, the 26 th term of the AP is 105.