Question

The 10th term of an A.P is 25 and the 18th term is 41. Find the

14th term of this A.P.


explain it step by step. ​

Answer 1

Given:-

• $a_{10} = 25$
• $a_{18} = 41$

To Find:-

• $a _{14}$

Solution:-

We Know that

$a _{n} = a + (n - 1)d$

So,

$a_{10} = a + (10 - 1)d = 25$

i.e,

$= > a_{10} =a + 9d = 25..(1)$

And Similarly,

$a_{18} = 41= a + (18 - 1)d$

$= > a_{18} = a + 17d = 41..(2)$

Now Subtracting equation (1) and (2)

$a + 17d - a - 9d = 41 - 25$

$= > 8d = 16$

$= > d = \frac{16}{8}$

$= > d = 2$

Now Substituting the value of d in Equation (1).

$a + 9d = 25$

$= > a + 9 \times 2 = 25$

$= > a = 18 = 25$

$= > a = 25 - 18$

$= > a = 7$

∴

• a=7
• d= 2

Now, we will Find $a _{14}$

So,

$a _{14} = 7 + 13 \times 2$

$= > a _{14} = 7 + 26$

$= > a _{14} = 33$

∴The 14th term of the A.P Is 33.

Answer 2

Answer:

Given:-

a_{10} = 25a

10

=25

a_{18} = 41a

18

=41

To Find:-

a _{14}a

14

Solution:-

We Know that

a _{n} = a + (n - 1)da

n

=a+(n−1)d

So,

a_{10} = a + (10 - 1)d = 25a

10

=a+(10−1)d=25

i.e,

= > a_{10} =a + 9d = 25..(1)=>a

10

=a+9d=25..(1)

And Similarly,

a_{18} = 41= a + (18 - 1)da

18

=41=a+(18−1)d

= > a_{18} = a + 17d = 41..(2)=>a

18

=a+17d=41..(2)

Now Subtracting equation (1) and (2)

a + 17d - a - 9d = 41 - 25a+17d−a−9d=41−25

= > 8d = 16=>8d=16

= > d = \frac{16}{8}=>d=

8

16

= > d = 2=>d=2

Now Substituting the value of d in Equation (1).

a + 9d = 25a+9d=25

= > a + 9 \times 2 = 25=>a+9×2=25

= > a = 18 = 25=>a=18=25

= > a = 25 - 18=>a=25−18

= > a = 7=>a=7

∴

a=7

d= 2

Now, we will Find a _{14}a

14

So,

a _{14} = 7 + 13 \times 2a

14

=7+13×2

= > a _{14} = 7 + 26=>a

14

=7+26

= > a _{14} = 33=>a

14

=33

∴The 14th term of the A.P Is 33.

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