Given the sum of the first 5 terms of an AP as 90 and the first 50 terms as 4275, what is the first term of the series?  

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giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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We'll apply the formula for the sum of n terms of an A.P.:

Sn = (a1+an)*n/2

We know that the sum of the first 5 terms is 90.

S5 = 90

90 = (a1+a5)*5/2

2*90 = (a1+a5)*5

We'll divide by 5 both sideS:

2*18 = a1 + a5

We'll write the formula for the general term of an A.P.

an = a1 + (n-1)*d

The common difference is d.

a5 = a1 + (5-1)d

a1 + a5 = 36 (1)

a5 = a1 + 4d (2)

We'll substitute (2) in (1):

2a1 + 4d = 36

We'll divide by 2:

a1 + 2d = 18 (3)

We also know that S50 = 4275.

S50 = (a1+a50)*50/2

4275 = (a1+a50)*25

We'll divide by 25 both sides:

a1 + a50 = 171 (4)

a50 = a1 + 49d (5)

We'll substitute (5) in (4):

2a1 + 49d  =171 (6)

We'll form the system from the equation (3) and (6):

a1 + 2d = 18 (3)

2a1 + 49d  =171 (6)

We'll multiply (3) by -2:

-2a1 - 4d = -36 (7)

We'll add (7) to (6):

-2a1 - 4d + 2a1 + 49d = -36 + 171

We'll eliminate and combine like terms:

45d = 135

We'll divide by 45:

d = 3

The common difference is 3.

We'll substitute d in (3):

a1 + 2d = 18

a1 + 6 = 18

We'll subtract 6 both sides:

a1 = 12

The first term of the A.P., whose common difference is 3 and sum of 5 first terms is 90, is a1 = 12.

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

We know that the sum of the n terms of an AP is given by:

Sn = {2a1+(n-1)d}n/2, where a1  is the first term, d is the common diffrerence between the successive terms.

Therefore sum of the 5 terms= S5 = {2a1+(5-1)d}5/2 = 90. Simplify this equation:

(2a1 +4d )5 = 90*2.

10a1 +20d = 180 .

Divide by 10:

a1+2d = 18...........(1) .

Also given the sum of first  50 terms = 4275.  So by formula, S50 = {2a1 +(50-1)d}50/2 =4275. Simplify this equation:

(2a1 +49d) = 4275/25.

2a1 +49d1 = 171..........(2).

Therefore  eq(2) - 2*eq(1) goves:

(2a1+49d)-2(a1+2d) = 171- 2*18.

45d = 171-18 =  135

.d = 135/45 = 3.

So substituting d = 3 in (1), we get: a1+2*3 = 18. So a1 = 18-6 = 12.

Therefore the first term of the AP  = a1 = 12.

william1941's profile pic

william1941 | College Teacher | (Level 3) Valedictorian

Posted on

An AP is a series which has a common difference between subsequent terms. The sum of the first n terms of an AP is given by the relation Sn = (n/2)*[2a + (n-1) d], where a is the first term of the AP and d is the common difference.

Now, it is given that the sum of the first 5 terms is 90

=> (5/2)*[2a + (5-1) d] = 90

=> (5/2)*(2a + 4d) = 90

=> 10a + 20d = 180

=> a + 2d = 18… (1)

Also, the sum of the first 50 terms is 4275.

=> (50/2)*[2a + (50-1) d] = 4275

=> 25*(2a +49d) = 4275

=> 50a + 1225d = 4275 … (2)

1225*(1) – 2*(2)

=> 1225a + 2450d – 100a – 2450d = 22050- 8550

=> 1125a = 13500

=> a= 13500/ 1125

=> a = 12

Therefore the AP starts with the term 12.

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