Question

find the 16th term of an Arithmetic progression ( A. p ) 7 , 11 , 15 , 19 .......

Find the sum of 1st 6th term..
Answer this fast please....​

Answer 1

Answer:

The given AP is 7 , 11 , 15 , 19 .......

So the first term = a = 7

Common Difference = d = 11 - 7 =4

By the required criteria n= 6

So the required sum of first 6th term

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

= 6/2 × [ 2 × 7 + ( 6 - 1) ×4]

= 3 × ( 14 + 20)

= 3 × 34

= 102

Answer 2

$\bf{\underline{Question:-}}$

find the 16th term of an Arithmetic progression ( A. p ) 7 , 11 , 15 , 19 .......

Find the sum of 1st 6th term..

$\bf{\underline{FORMULA:-}}$

★ $\bf\large a_n = a + (n-1) d$

★ $\bf\large S_n = \frac{n}{2}[2a+ (n-1)d$

$\bf {\underline{Given:-}}$

• First term ( a ) = 7
• Common difference ( d ) = 4
• Sum of 16th term = ?
• Sum of 1st 6th term = ?

$\bf{\underline{Solution:-}}$

★ Finding the sum of 16th term.

→ $\bf a_n = a+ (n-1) d$

→ $\bf a_{16} = 7 + (16 - 1) 4$

→ $\bf a_{16} = 7 + 15 × 4$

→$\bf a_{16} = 67$

★Now calculating sum of first 6th term.

→ $\bf S_n = \frac{n}{2}[2a+ (n-1)d$

→ $\bf S_6 = \frac{6}{2}[ 2× 7 + (6 - 1) 4]$

→ $\bf S_6 = 3 [ 14 + 5 × 4]$

→ $\bf S_6 = 3 [ 14 + 20]$

→ $\bf S_6 = 3 × 34$

→ $\bf S_6 = 102$

$\bf{\underline{Hence:-}}$

• Sum of 16th term = 67
• The sum of First 6th term = 102