Question

The 17th term of an A.P. is 5 more than twice its 8th term. If the 11th term of the A.P is 43, find the 10th term

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Answer 1

Answer:

38

Step-by-step explanation:

8th term is t8.

10th term is t10.

11th term is t11.

17th term is t17.

t8 = a + 7d

t10 = a + 9d

t11 = a + 10d

t17 = a + 16d

The 17th term of an A.P. is 5 more than twice its 8th term,

t17 = 5 + t8

t17 - t8 = 5

a + 16d - ( a + 7d ) = 5

a + 16d - a - 7d = 5

a - a + 16d - 7d = 5

9d = 5 ------> 1

d = 5 / 9 -----> 2.

t11 = 43

a + 10d = 43

Substitute 2.

a + 10 ( 5 / 9 ) = 43

a + ( 50 / 9 ) = 43

( 9a / 9 ) + ( 50 / 9 ) = 43

( 9a + 50 ) / 9 = 43

9a + 50 = 43 × 9

9a + 50 = 387

9a = 337 --------> 3.

a = 337 / 9

t10 = a + 9d

Substitute 1 and 3.

t10 = ( 337 / 9 ) + 5

= ( 337 / 9 ) + ( 5 / 9 )

= ( 337 + 5 ) / 9

= 342 / 9

t10 = 38

Therefore, 10th term ( t10 ) = 38.