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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.
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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.
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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.....
Given :-
To Find :-
Formula used :-
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.