Question

If for an AP. t10= 25 and t18= 45. Find the value of t14=?​

Answer 1

Answer:

$\red\bigstar$ The value of t₁₄ = 35.

SOLUTION

We know that ,

tn = a + (n - 1)d

Therefore,

t₁₀ = a + (10 - 1)d

⇒ t₁₀ = a + 9d

∴  a + 9d = 25 $\longrightarrow$(1)

Now Similarly,

t₁₈ = a + (18 - 1)d

⇒ t₁₈ = a + 17d

∴ a + 17d = 45 $\longrightarrow$ (2)

Next we must subtract (1) from (2)

    a + 17d = 45

   -a - 9d = -25  

         8d  = 20

$\implies$  8d = 20

$\implies \sf d = \dfrac{20}{8} = \dfrac{5}{2}$

Now, let us substitute d = 5/2 in (1).

$\implies \sf a + 9*\dfrac{5}{2} = 25$

$\implies \sf a + \dfrac{45}{2} = 25$

$\implies \sf \dfrac{2a + 45}{2} = 25$

$\implies$ 2a + 45 = 25 × 2

$\implies$ 2a + 45 = 50

$\implies$ 2a = 50 - 45

$\implies$ 2a = 5

$\implies \sf a = \dfrac{5}{2}$

Now , ∵ We have the value of both d and a we can know obtain value of t₁₄.

$\implies$ t₁₄ = a + (14 - 1)d

$\implies$  t₁₄ = a + 13d

Here,

$\sf a = \dfrac{5}{2}$

$\sf d = \dfrac{5}{2}$

$\implies \sf t_{14} = \dfrac{5}{2} + 13 ( \dfrac{5}{2})$

Now, let us take 5/2 out as common.

$\implies \sf t_{14} = \dfrac{5}{2} *( 13 + 1)$

$\implies \sf t_{14} = \dfrac{5}{2} *14$

$\implies \sf t_{14} = \dfrac{70}{2}$

$\implies$ t₁₄ = 35

✳ t₁₄ = 35

Answer 2

Answer:

★ The value of t₁₄ = 35.

SOLUTION

We know that ,

tn = a + (n - 1)d

Therefore,

t₁₀ = a + (10 - 1)d

⇒ t₁₀ = a + 9d

∴ a + 9d = 25 \longrightarrow⟶ (1)

Now Similarly,

t₁₈ = a + (18 - 1)d

⇒ t₁₈ = a + 17d

∴ a + 17d = 45 \longrightarrow⟶ (2)

Next we must subtract (1) from (2)

a + 17d = 45

-a - 9d = -25

8d = 20

\implies⟹ 8d = 20

\implies \sf d = \dfrac{20}{8} = \dfrac{5}{2}⟹d=

8

20

=

2

5

Now, let us substitute d = 5/2 in (1).

\implies \sf a + 9*\dfrac{5}{2} = 25⟹a+9∗

2

5

=25

\implies \sf a + \dfrac{45}{2} = 25⟹a+

2

45

=25

\implies \sf \dfrac{2a + 45}{2} = 25⟹

2

2a+45

=25

\implies⟹ 2a + 45 = 25 × 2

\implies⟹ 2a + 45 = 50

\implies⟹ 2a = 50 - 45

\implies⟹ 2a = 5

\implies \sf a = \dfrac{5}{2}⟹a=

2

5

Now , ∵ We have the value of both d and a we can know obtain value of t₁₄.

\implies⟹ t₁₄ = a + (14 - 1)d

\implies⟹ t₁₄ = a + 13d

Here,

\sf a = \dfrac{5}{2}a=

2

5

\sf d = \dfrac{5}{2}d=

2

5

\implies \sf t_{14} = \dfrac{5}{2} + 13 ( \dfrac{5}{2})⟹t

14

=

2

5

+13(

2

5

)

Now, let us take 5/2 out as common.

\implies \sf t_{14} = \dfrac{5}{2} *( 13 + 1)⟹t

14

=

2

5

∗(13+1)

\implies \sf t_{14} = \dfrac{5}{2} *14⟹t

14

=

2

5

∗14

\implies \sf t_{14} = \dfrac{70}{2}⟹t

14

=

2

70

\implies⟹ t₁₄ = 35

✳ t₁₄ = 35