Question

the 13th term of an ap is 2 times its 10 th term and the third term of 6 more than two times its six times finds the first 3 terms​

Answer 1

Answer:

The first three terms of AP is $t_1$ = 18 ,  $t_2$ = 15 ,  $t_3$  = 12   .

Step-by-step explanation:

Given as :

The 13th term of an AP is 2 times its 10th term

Since nth  term of an AP is given as

$t_n$ = a + (n - 1) d

where a is the first term

and d is the common difference between terms

So, $t_13$ = a + (13 - 1) d

Or,  $t_13$ = a + 12 d                 ......A

And

$t_10$ = a + (10 - 1) d

Or, $t_10$ = a + 9 d                 ..............B

According to question

$t_13$ = two times  $t_10$

i.e  $t_13$ = 2 × $t_10$

From eq A and eq B

a + 12 d = 2 × (a + 9 d)

or, a + 12 d = 2 a + 18 d

Or, 2 a + 18 d - a - 12 d = 0

i.e a + 6 d = 0                        ..........1

Again

The third term is 6 more than two times its six times

So, $t_6$ = a + (6 - 1) d

i.e $t_6$ = a + 5 d                    ...............C

And

$t_3$  = a + (3 - 1) d

i.e $t_3$  = a + 2 d                    ........D

From eq C and eq D

$t_3$  = 6 + 2 ×  $t_6$

Or, a + 2 d = 6 + 2 × (a + 5 d)

Or, a + 2 d = 6 + 2 a + 10 d

Or, 2 a - a + 10 d - 2 d + 6 = 0

Or, a + 8 d = - 6                         ..........2

Solving eq 1 and eq 2 we get

(a + 8 d) - (a + 6 d) = - 6 - 0

Or, (a - a) + (8 d - 6 d) = - 6

Or, 0 + 2 d = - 6

i.e d = $\dfrac{-6}{2}$

Or, d = - 3

Put the value of d in eq 1

∵ a + 6 d = 0

Or, a + 6 (- 3) = 0

Or, a - 18 = 0

i.e a = 18

So, The third term =  $t_3$  = a + 2 d  

i.e  $t_3$  = 18 + 2 × ( - 3)

Or, $t_3$  = 18 - 6

∴   $t_3$  = 12

And , The second term = $t_2$ = a + (2 - 1) d

i.e  $t_2$ = a + d

Or,  $t_2$ = 18 + (- 3)

∴,  $t_2$ = 15

And , The first term = $t_1$ = a + (1 - 1) d

i.e  $t_1$ = 18 + 0 × (-3)

∴ $t_1$ = 18

Hence, The first three terms of AP is $t_1$ = 18 ,  $t_2$ = 15 ,  $t_3$  = 12   . Answer

Answer 2

Step-by-step explanation:

please except me as brainlist