Question

The sum of first six terms of an A.P. is 69 and the
sum of last three terms of the same A.P. is 273.
If the first term of the A.P. is 4, find the number of
terms in that ​A.P​

Answer 1

Hey Dear,

◆ Answer -

31 terms

● Explanation -

Consider an AP with first term a, common difference d & last term l.

Information given is -

a = 4

S6 = 69

Sum of n terms in AP is given by -

Sn = n/2 [2a + (n-1)d]

S6 = 6/2 [2×4 + (6-1)d]

69 / 3 = 8 + 5d

5d = 23 - 8

d = 15 / 5

d = 3

Last term in AP is calculated by -

l = S/3 + d

l = 273/3 + 3

l = 91 + 3

l = 94

But, we know that -

l = a + (n-1)d

94 = 4 + (n-1)3

94 - 4 = 3n - 3

3n = 93

n = 31

Therefore, Given AP consists of 31 terms.

Thanks dear..

Answer 2

Answer:

$\huge{\boxed{\boxed{\sf{n=31}}}}$

Step-by-step explanation:

$\huge{\boxed{\boxed{\underline{\sf{SOLUTION-}}}}}$

□a=4

□S6=69

□SL=273 (sum of last three terms)

To find:

○A.P=?

○n=?

$S6 = 69 \\ = ) \frac{n}{2} (2a +( n - 1) \times d) = 69 \\ = ) \frac{6}{2} (2 \times 4 + 5 \times d) = 69 \\ = )3(8 + 5d) = 69 \\ = )8 + 5d = \frac{69}{3} \\ = )5d = 23 - 8 \\ = )d = \frac{15}{5} \\ = )d = 3 \\ again \\ L = \frac{SL}{3} + d \\ = ) \frac{273}{3} + d \\ = )91 + 3 \\ = )94 \\ last \: term(L) \: of \: this \: ap \: is \: 94 \\ so \: we \: find \: number \: of \: terms \\ L= a + (n - 1) \times d \\ = ) 94 = 4 + (n - 1) \times 3 \\ = )94 - 4=( n - 1) \times 3 \\ = ) \frac{90}{3} = n - 1 \\ = ) n = 30 + 1 \\ = )n = 31$

☆n=31

☆A.P=4,7,10,13........94

$\huge{\boxed{\boxed{\sf{n=31}}}}$