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:
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.
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