Question

In an arithmetic sequence, 7th term is 40, 13th term is 80.
A) Find 10th Term.
B) Find the sum of first 19 terms of this sequence.
C) In an arithmetic sequence, 7th term is 42 and 13th term is 82. Then what is the sum of first 19 terms?
Plz give me the answer.. ​

Answer 1

If any mistake is there please inform me

Answer 2

Answer:

A) 10th Term is 60.

B) The sum of first 19 terms of this sequence is 1140.

C) The sum of first 19 terms is 1197

Step-by-step explanation:

A) We know,nth term in AP series

$a_n = a + (n - 1) \times d$

Where a is first term and d is common difference of the series.Given,7th term is 40 and 13th term is 80. So,

$a_7 = a + (7 - 1) \times d \\ a + 6d = 40....(1)$

and

$a_{13} = a + (13- 1) \times d \\ a + 12d = 80....(2)$

We are subtracting equation (1) from equation (2),

$a + 12d - a - 6d = 80 - 40 \\ 6d = 40 \\ d = \frac{40}{6} \\ d = \frac{20}{3}$

From equation (1),

$a + 6 \times \frac{20}{3} = 40 \\ a + 40 = 40 \\ a = 0$

Now we want to find 10th term.So,

$[tex]a_{10}= 0 + (10- 1) \times \frac{20}{3} \\ = 0 + 9 \times \frac{20}{3} \\ = 0 + 3 \times 20 \\ = 60$[/tex]

B) We know sum of n terms in AP series,

$\frac{n}{2} [2a + (n - 1)\times d$

So, sum of 19 terms in AP series

$= \frac{19}{2} [2 \times 0 + (19 - 1) \times \frac{20}{3} ] \\ = \frac{19}{2} \times 18 \times \frac{20}{3} \\ = 19 \times 3 \times 20 \\ = 1140$

C)Again it is also given,a another AP series which has 7th term is 42 and 13th term is 82.

$a_7 = a + (7 - 1) \times d \\ a + 6d = 42....(3)$

and

$a_{13} = a + (13 - 1) \times d \\ a + 12d = 82....(4)$

From equation (3) and (4),

$6d = 40 \\ d = \frac{40}{6} \\ d = \frac{20}{3}$

and from equation (3),

$a + 6 \times \frac{20}{3} = 42 \\ a + 40 = 42 \\ a = 2$

So, required sum of first 19 terms

$= \frac{19}{2} [2 \times 2 + (19 - 1)\times \frac{20}{3} ] \\ = \frac{19}{2} \times 126 \\ = 1197$

Know more about Arithmetic progression:

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