Question

In an arithmetic sequence t5 = 12, t10 = 26 find t15.

Answer 1

Answer:

$t_{15} = 40$

Given----->

$t_{5} = 12 \\ \\ t_{10} = 26$

to find:-

value of t 15

$as \: we \: know \\ t_{n} = a + (n - 1)d \\ \\ t_{5} = a + (5 - 1)d \\ 12 = a+ 4d \: \: .............(1) \\ \\ t_{10} = a + (10 - 1)d \\ 26 = a + 9d \: \: ................(2) \\ \\ now \: we \: will \: solve \: the \: \\ as \: linear \: equation \\ \\ \: substract \: equation \: 1 \: and \:2 \\ \\ \: \: a + 4d = 12 \\ - a + 9d = 26 \\ - - - - - - - - \\ - 5d = - 14 \\ d = \frac{14}{5} \: \\ \\ now \: put \: d = \frac{14}{5} in \: equation \: 1 \\ \\ a + 4d = 12 \\ \\ a + 4 \times \frac{14}{5} = 12 \\ \\ \frac{5a + 56}{5} = 12 \\ \\ 5a + 56 = 60 \\ \\ 5a = 60 - 56 \\ \\ 5a = 4 \\ \\ a = \frac{4}{5} \\ \\ now \: for \: t_{15} \\ \\ t_{n} = a + (n - 1)d \\ put \: a = \frac{4}{5} and \: d \: = \frac{14}{5} \\ \\ t_{15} = \frac{4}{5} + (15 - 1) \times \frac{14}{5} \\ \\ t_{15} = \frac{4}{5} + 14 \times \frac{14}{5} \\ \\ \frac{4}{5} + \frac{196}{5} \\ \\ t_{15} = \frac{196 + 4}{5} \\ \\ t_{15} = \frac{200}{5} \\ \\ t_{15} = 40$

Answer 2

$t_{15} =40$

Step-by-step explanation:

Let first term = a and common difference = d

Given,

$t_5 = 12$ and  $t_{10}=26$

To find, the value of $t_{15}=?$

We know that,

The nth term of an arithmetic progression,

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

∴ $t_{5} =a+(5-1)d$

⇒ $a+4d=12$       .....(1)

Also,

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

⇒ $a+9d=26$    .....(2)

Subtracting (2) from (1), we get

9d - 4d = 26 - 12

⇒ 5d = 14

⇒ $d=\dfrac{14}{5}$

Put $d=\dfrac{14}{5}$ in (1), we get

$a+4(\dfrac{14}{5} )=12$

⇒ $a=12-\dfrac{56}{5} =\dfrac{60-56}{5}$

⇒ $a=\dfrac{4}{5}$

The 15th term of an AP

∴ $t_{15} =a+14d$

⇒$t_{15} =\dfrac{4}{5} +14(\dfrac{14}{5})$

⇒ $t_{15} =\dfrac{196+4}{5}=\dfrac{200}{5}$

⇒ $t_{15} =40$

Hence, $t_{15} =40$