Question

.
The thirteenth term of an A.P is two times its tenth term and the third term
is 6 more than two times its sixth term. Find the first three terms.​

Answer 1

Answer:

a 13 =a+12d=2×a10

a+12d=2(a+9d)

-a-6d=0.........(1)

a3= a+2d= 2(a+5d)+6

-a-8d=6..........(2)

By(1)-(2):

-2d=6

d= -3

a= 18

A.P: 18,15,12.....

Answer 2

Given :-

• 13th term of AP = 2 * 10th term.
• 3rd Term = 2 * 6th term + 6

To Find :-

• First Three Terms of AP ?

Formula used :-

• nth Term of AP = a + (n - 1)d , where a = First term , d = common Difference.

Solution :-

13th term of AP = 2 * 10th term.

So,

→ a + (13 - 1)d = 2 * [ a + (10 - 1)d ]

→ a + 12d = 2(a + 9d)

→ a + 12d = 2a + 18d

→ 2a - a = 12d - 18d

→ a = (-6)d. ------------- Equation (1)

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Now,

3rd Term = 2 * 6th term + 6

→ a + (3 - 1)d = 2 (a + (6 - 1)d + 6

→ a + 2d = 2(a + 5d) + 6

→ a + 2d = 2a + 10d + 6

→ 2a - a = 2d - 10d - 6

→ a = - 8d - 6 ---------------- Equation (2).

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Putting Value of Equation (1) in Equation (2) now, we get,

→ -6d = -8d - 6

→ -6d + 8d = (-6)

→ 2d = (-6)

→ d = (-3).

Putting This value in Equation (1) now, we get,

→ a = (-6) * (-3) = 18 .

Hence,

→ First Term of AP = a = 18.

→ Second Term = a + d = 18 - 3 = 15

→ Third Term = a + 2d = 18 - 2*3 = 18 - 6 = 12.