The sum of the fourth and twelfth term of an arithmetic

OA C

Sum of the AP = avg of AP x no of terms
Avg of the AP can be found by addition of (first term + last term) /2 or (second +second last term)/2 or (third +third last term)/2.... try this out its a property of any AP....here as we need to find sum of first 15 terms...let us assume AP of 15 terms....

for this AP..going by above logic..... (1st +15th term) = (4th + 12 th term) = 20
hence Avg of AP = 10

Sum of the AP = 10 x 15 = 150

For any Arithmetic Progression, the n'th term can be written as (a + (n-1)d), where 'a' = first term, 'd' = Common difference between two terms, and 'n' = number of terms.
e.g. 3rd term = a+2d, 1st term = a + 0d = a

The sum of the fourth and twelfth term of an arithmetic progression is 20.
4th Term + 12th Term = 20 ----------> (a+3d) + (a+11d) = 20 ----------> 2a + 14d = 20


What is the sum of the first 15 terms of the arithmetic progression?

Sum of n terms of a arithmetic progression is given by \(Sum = \frac{n(a+l)}{2}\), where 'n' = number of terms, 'a' = first term, and 'l' = last term

\(Sum of first 15 terms = \frac{15(1st term+15th term)}{2}\) ----------> \(\frac{15(a+(a+14d))}{2}\) ------------> \(\frac{15(2a+14d)}{2}\) ---------> \(\frac{(15)(20)}{2}\) -------------> \(150\)

IMO C...

n th term of A.P. is given by a+(n-1)d
4 th term = a+3d
12 th term = a+11d

Given a+3d+a+11d=20 --> 2a+14d=20 --> a+7d = 10

Sum of n term of A.P = n/2[2a+(n-1)d]
subsitiuing n = 15 ...we get 15/2[ 2a + 14d] = 15 [a+7d] = 15*10 = 150...

Formula for Finding the nth term of an A.P. = Tn = a + (n-1) d

Using this we can calculate the the fourth term and the 12th term of this AP ..

T4 = a + 3d
T12 = a+11 d

We know that T4 + T12 = 20 a

Therefore ; a+3d + a + 11d = 20 ----> 2a + 14 d...

The question asks us to find the sum of the first 15 terms ..

Therefore using the formula for the Sum of n terms of an AP we get :

S15 = n/2 [ 2a + (15-1)d] [/b] ----> n/2 [ 2a + 14 d]

We have calculated the value for 2a + 14 d to be 20 , therefore we have S 15 = 15/2 (20) = 150 (C)
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I'm sorry if this is not germane to the post but in what particular math class or type of math do you learn these formulas such as the sum of terms in a series? I have done elementary-calculus and am now in a college level statistics course. I have not seen any of these series formulas.

let a be the first term n be the number of terms and d be the difference

a+3d +a +11d = 20 or
2a +14d = 20

sum of 15 terms = n/2{2a + (n-1)d}
= 15/2(2a +14d)
substituting we get

15*20/2 =150.

option C

Let fourth term be t4 and 12th term be t12.

Now t12 can be written as : t12 = t4 + 8d; ---- (Equation I)

Given : t4 + t12 = 20 ---- (Equation II)

Substituting value of t12 in Equation II, we get :

t4 + t4 + 8d = 20

=> t4 + 4d = 10

Which is nothing but t8. (Since t4 + 4d = t8)

We now the value of t8;

Also, 15 terms in AP can also be written as :

a - 7d, a - 6d, a - 5d, a - 4d, a - 3d, a - 2d, a - d, a, a + d, a + 2d, a + 3d, a + 4d, a + 5d, a + 6d, a + 7d ; ---- (Equation III)

If you notice, a is nothing but the 8th term in above series.

Taking sum of all the terms in equation III, we get,

S15 = 15a = 15 * t8 = 15 * 10 = 150

which is the required answer. Option C

let x=8th term
d=difference between terms
(x-4d)+(x+4d)=20➡
x=10
because 10 is the median of the first 15 terms,
their sum will be 15*10=150
C

Bunuel,

Which approach is more preferable while solving such questions?
1.Sum of Ap = Average of Ap * No. of terms.

or

2. Converting them like
4 th term = a+3d
12 th term = a+11d
andthen adding them up?

I am asking this because I am not sure about the first approach.

T8=(T12+T4)/2=20/2=10
Sum of Series = Middle term*Number of Terms
10*15
=150

The problem is simple S4 + S12=20
a+3d +a +11d=20
a+a15=20
S15=n/2(a+a15)=15/2(20)=150

Term = T
Common difference = D

T4+T12=20
T4=T1+3D
T12= T1+11D

2T1+14D=20 (1)

Sum of first 15 terms = 15*(T1+T15)/2
T15=T1+14D

Sum of first 15 terms = 15*(2T1+14D)/2 (2)

(1) & (2)
Answer is 150

We know the sum of 15 terms = n/2{2a + (n-1)d}. So the sum should be divisible by 15, with this we can eliminate option d.
next we know two of the terms add up to 20.
So the sum - 20 = should be the sum of 13 terms. Which means the option should be divisible by 13.

A. 300 - 300-20= 280, not divisible
B. 120 -120-20 =100/13 -ND
C. 150 150-20 = 130 ANSWER
D. 170 170-30 - 150 ND
E. 270 270-20 = 250 ND