find
an AP whose 6th term is 14 and the sum of its 7th term and 15th term is 112
Answers
EXPLANATION.
6th term is 14.
Sum of it's 7th term and 15th term is 112.
As we know that,
General terms of an ap.
⇒ Tₙ = a + (n - 1)d.
Using this formula in this question, we get.
⇒ T₆ = 14.
⇒ a + (6 - 1)d = 14.
⇒ a + 5d = 14. - - - - - (1).
⇒ T₇ + T₁₅ = 112.
⇒ [a + (7 - 1)d] + [a + (15 - 1)d] = 112.
⇒ a + 6d + a + 14d = 112.
⇒ 2a + 20d = 112.
⇒ a + 10d = 56. - - - - - (2).
From equation (1) and (2), we get.
Subtract both the equation, we get.
⇒ a + 5d = 14. - - - - - (1).
⇒ a + 10d = 56. - - - - - (2).
⇒ - - -
We get,
⇒ 5d - 10d = 14 - 56.
⇒ - 5d = - 42.
⇒ 5d = 42.
⇒ d = 42/5.
Put the value of d = 42/5 in equation (1), we get.
⇒ a + 5d = 14. - - - - - (1).
⇒ a + 5 x (42/5) = 14.
⇒ a + 42 = 14.
⇒ a = 14 - 42.
⇒ a = - 28.
First term = a : - 28.
Common difference = d : 42/5.
As we know that,
Arithmetic Progression (A.P) : If a is the first term and d is the common difference then A.P can be written as,
⇒ a + (a + d) + (a + 2d) + . . . . .
Now, we can write series as,
⇒ - 28, [- 28 + (42/5)], [- 28 + 2(42/5)], . . . . .
⇒ - 28, [(-140 + 42)/(5)], [(-140 + 84)/(5)], . . . . .
⇒ - 28, [-98/5], [-56/5], . . . . .
∴ A.P's are : - 28, (-98/5), (-56/5), . . . . .