Question

the first and the last term of an AP are 5 and 45 .if the sum of all its terms is 400 .find the comman difference​

Answer 1

Answer:

last term = ,5

first term = 45

Sn of n term =400

Sn = n/2(a+l)

400= n/2(5+45)

400 = 25n

n = 16

l = a+(n-1)d

45 = 5+( 16-1)d

d= 40/15

d=8/3

Answer 2

Question:-

The first and the last term of an AP are 5 and 45.if the sum of all its terms is 400 .find the comman difference.

Solution:-

Given,

• First term of an A.P is 5 and the last term of an A.P is 45 and the sum of the terms is 400.

➣ First term (a) = 5

➣ Last term (l) = 45

➣ Sum of all the terms = 400

➣ common difference (d) = ?

➣ Sn = n/2 [ first term + last term ]

➣ 400 = n/2 [ 5 + 45 ]

➣ 400/50 = n/2

➣ n = (400 × 2)/50

➣ n = 800/50

➣ n = 16

∴ 45 is the 16th term and the last term of is 16th term.

Now,

We have to find the common difference.

➣ tn = a + (n - 1)d

➣ 45 = 5 + (16 - 1)d

➣ 45 = 5 + 15d

➣ 40 = 15d

➣ 40/15 = d

➣ d = 8/3

${\boxed{∴ The \: common \: di</strong><strong>f</strong><strong>f</strong><strong>e</strong><strong>rence \: of \: A.P \: is \: \frac{8}{3} .}}$