Question

1) Find the sum of the first 14 terms of an A.P. 3, 15, 27, 39, …


2)Find the 7th term from the end of the arithmetic progression 7, 10, 13,…., 84

pls tell both answers fast pls they both are different questions from PROGRESSIONS chapter

Answer 1

Answer:

1) 1134

2) 166

Step-by-step explanation:

Question - 1 :

Given A.P. : 3 , 15 , 27 , 39 , ......

• first term, a = 3

• common difference is the difference between a term and the preceding term.

        Common difference, d = 15 - 3 = 12

we have to find the sum of first 14 terms.

 

⇒ Sum of first n terms is given by,

$\boxed{ \bf S_n= \dfrac{n}{2} [2a+(n-1)d]}}$

where

a is the first term

d is the common difference

Substitute n = 14 and the values of a and d,

     $\sf S_{14} =\dfrac{14}{2} [2(3)+(14-1)(12)] \\\\ S_{14} =7[6+13(12)] \\\\ S_{14}=7[6+156] \\\\ S_{14}=7[162] \\\\ S_{14}=1134$

∴ Sum of first 14 terms = 1134

Question - 2 :

Correct Question : Find the 7th term from the end of the arithmetic progression 7, 10, 13,…., 184

Common difference = 10 - 7 = 3

To find the nth term from the end, reverse the A.P.

 184 , ..... , 13 , 10 , 7

Then

• first term, a = 184
• common difference, d = -3 ( since we reversed the A.P. )

nth term of an A.P. is given by,

$\boxed{\bf a_n=a+(n-1)d]}$

where

a is the first term

d is the common difference

Substitute n = 7 , a = 184 and d = -3,

➙ a₇ = 184 + (7 - 1)(-3)

➙ a₇ = 184 + 6(-3)

➙ a₇ = 184 - 18

➙ a₇ = 166

∴ The 7th term from the end of the given A.P. is 166