Find an
a.P whose 6th term is 14 and sum of its 7th term and 15th term is 112.
Answers
A6= 14
A1+5d=14
A7+a15= 112
A1+6d+a1+14d=112
2a1+20d=112
A1+10d=56
A1+5d+5d=56
5d+14=56
5d= 12
D= 12/5
A1+5d=14
A1=2
Find the other terms using the values I just found out
Please mark my answer as the brainliest
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), . . . . .