Question

if the sum of n successive odd
natural humbers starting from 3 is 48, find the value of n​

Answer 1

a = 3

last term = 47

d = 2

Last term = a + (n-1)d

47 = 3 + 2n - 2

47 - 1 = 2n

2n = 46

n = 46/2 or 23 terms

Answer 2

$\bold{\underline{\underline{\huge{\sf{ANSWER\::}}}}}$

Given:

If the sum of n successive odd natural numbers starting from 3 is 48.

To find:

The vaue of n.

Explanation:

We have,

Odd natural number starting from 3,5,7...........nth term.

We know that sum of the A.P.;

Sn= $\frac{n}{2}$ [2a+(n-1)d

Given,

• First term of an A.P.,a= 3
• Common difference,d=2
• Sn term = 48

Therefore,

→ Sn=$\frac{n}{2}$ [2×3+(n-1)2]

→ 48= $\frac{n}{2}$ [6+(n-1)2]

→ 96= n[6 +2n-2]

→ 96= 6n +2n² -2n

→ 96= 2n² +4n

→ 2n² +4n -96=0

→ n²+ 2n -48= 0

[Factorise]

→ n²+ 8n-6n-48=0

→ n(n+8) -6(n+8)=0

→ (n+8)(n-6)=0

→ n+8=0    or     n-6=0

→ n= -8      0r     n= 6

n= -8 is not acceptable value because negative value.

∴ n= 6

Thus,

The value of n is 6.