sovle question11 of mathematics
Answers
Answer:
2nd term of the AP is - 8
rth term of the AP is 5r - 18
Step-by-step explanation:
Given,
6th term = 12
8th term = 22
From the identities of arithmetic progressions,
nth term = a + ( n - 1 )d
∴ We can write 6th term of the AP is a + ( 6 - 1 )d i.e. a + 5d and 8th term of the AP as a + ( 8 - 1 )d i.e. a + 7d .
Then, a + 5d = 6th term = 12
a + 7d = 8th term = 22
⇒ a + 5d = 12
⇒ a = 12 - 5d ...( i )
Substitute the value of a from ( i ) in 8th term.
⇒ a + 7d = 22
⇒ 12 - 5d + 7d = 22
⇒ 12 + 2d = 22
⇒ 6 + d = 11
⇒ d = 11 - 6
⇒ d = 5
Now, substitute the value of d in ( i ),
⇒ a = 12 - 5d
⇒ a = 12 - 5( 5 )
⇒ a = 12 - 25
⇒ a = - 13
∴ 2nd term of the AP = a + ( 2 - 1 )d
⇒ - 13 + ( 1 x 5 )
⇒ - 13 + 5
⇒ - 8
And, rth term of the AP = a + ( r - 1 )d
⇒ - 13 + ( r - 1 ) x 5
⇒ - 13 + 5r - 5
⇒ 5r - 18