Question

if 6 term of an ap is -10 and 10term -26 then find the 15term

Answer 1

Answer:

- 46

Step-by-step explanation:

From the properties of AP :

        nth term = a + ( n - 1 )d            { where a is the first term and d is the common difference between the terms }.

Let : first term of this AP be a and d be the common difference.

6th term is - 10:

⇒ a + ( 6 - 1 )d = - 10

⇒ a + 5d = - 10                 ...( 1 )

10th term is -26 :

⇒ a + ( 10 - 1 )d = - 26

⇒ a + 9d = - 26             ...( 2 )

⇒ ( 2 ) - ( 1 )

⇒ a + 9d - a - 5d = - 26 - (- 10 )

⇒ 4d = - 26 + 10

⇒ 4d = - 16

⇒ d = - 4

            Thus,

            a + 5d = - 10

            a + 5( - 4 ) = -10

            a = - 10 + 20

            a = 10

Hence 15 th term is :

⇒  a + 14d

⇒  10 + 14( - 4 )

⇒ 10 - 56

⇒ - 46

Answer 2

Given :

• 6 th term of an ap is -10.
• 10 th term of the ap is -26.

To Find :

• The 15th term, $\bold{t_{15}}$

Solution :

Let the first term of the ap be a.

Let the common difference of the ap be d.

We know the formula to find nth term of an AP,

$\large{\bold{\red{t_n\:=\:a\:+\:(n-1)d}}}$

6th term of AP :

Here,

• $\sf{t_n\:=\:-10}$
• $\sf{n=6}$

Block the values in the formula,

$\longrightarrow$ $\sf{-10\:=\:a+\:(6-1)d}$

$\longrightarrow$ $\sf{-10=a+(5)d}$

$\longrightarrow$ $\sf{-10=a+5d\:\:(1)}$

10th term of AP :

Here,

• $\sf{t_n\:=\:-26}$
• $\sf{n=10}$

Block in the values,

$\longrightarrow$ $\sf{-26=a+(10-1)d}$

$\longrightarrow$ $\sf{-26=a+(9)d}$

$\longrightarrow$ $\sf{-26=a+9d\:\:\:(2)}$

Now, substract equation (2) from (1),

$\longrightarrow$ $\sf{a+5d-(a+9d)=-10-(-26)}$

$\longrightarrow$ $\sf{a+5d-a-9d=-10+26}$

$\longrightarrow$ $\sf{5d-9d=16}$

$\longrightarrow$ $\sf{-4d=16}$

$\longrightarrow$ $\sf{d=\dfrac{16}{-4}}$

$\longrightarrow$ $\sf{d=-4}$

Substitute, d = 4 in equation (1)

$\longrightarrow$ $\sf{a+5d=-10}$

$\longrightarrow$ $\sf{a+5(-4)=-10}$

$\longrightarrow$ $\sf{a+(-20)=-10}$

$\longrightarrow$ $\sf{a-20=-10}$

$\longrightarrow$ $\sf{a=-10+20}$

$\longrightarrow$ $\sf{a=10}$

15th term of AP :

Using the formula of $\sf{t_n}$,

$\longrightarrow$ $\sf{t_{15}\:=\:a\:+\:(15-1)d}$

$\longrightarrow$ $\sf{t_{15}\:=\:10+(14)(-4)}$

$\longrightarrow$ $\sf{t_{15}\:=\:10\:+\:(-56)}$

$\longrightarrow$ $\sf{t_{15}\:=\:10-56}$

$\longrightarrow$ $\sf{t_{15}=-46}$

$\large{\boxed{\bold{15th\:term\:,t_{15}\:=\:-46}}}$