Question

For an AP T5=26 and T10=51 find T15.

Answer 1

Given, For an AP,

$\rm{}t_{ \tiny{5}} = 26 \\ \rm{}t_{ \tiny{10}} = 51$

$\bull \: \: \text{To find } : \: \green{ \rm{}t_{ \tiny{15}}}$

We know,

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

• where, a is the first term, d is common difference, and n is the number of terms in the AP

So,

$\rm{}t_{ \tiny{5}} = a + (5 - 1)d = 26 \\ = > a + 4d = 26 \: \: \: \: \: \: \: \: \: \: \: \: \cdot\cdot\cdot\cdot(1)$

And,

$\rm{}t_{ \tiny{10}} = a + (10 - 1)d = 51 \\ = > a + 9d = 51\: \: \: \: \: \: \: \: \: \: \: \: \cdot\cdot\cdot\cdot(2)$

Now, Subtracting Equation 2 from Equation 1

$= > a + 4d - (a + 9d) = 26 - 51 \\ = > \cancel{a} + 4d - \cancel{a} - 9d = - 25 \\ = > \: \: \: \: \: \: \cancel- 5d = \cancel - 25 \\ = > \: \: \: \: \: \:5d = 25 \\ \\ = > \frac{\cancel5d}{\cancel5} = \frac{25}{5} \\ \\ = > \: \: \: \: \: \:d = \green5$

So, Putting value of d in equation 1 ,

$= > a + 4d = 26 \\ = > a + 4(5) = 26 \\ = > a + 20 = 26 \\ = > \: \: \: \: \: a = 26 - 20 \\ = > \: \: \: \: \: a = \green6$

Thus,

$= > \rm{}t_{ \tiny{15}} = a + (15 - 1)d \\ = > \rm{}t_{ \tiny{15}} = 6 + 14(5) \\ = > \rm{}t_{ \tiny{15}} = 6 + 70 \: \: = \pink{76}$

Answer 2

your answer is perfectly correct