Question

The 10th and 18th terms of an A.P. are 41 and 73 respectively. Find 26th term.​

Answer 1

Answer:

$\huge{\boxed{\sf{ a26=105 }}}$

Step-by-step explanation:

$\huge{\boxed{\sf{ SOLUTION- }}}$

$a10 = 41 \\ a + 9d = 41 - - - - - (1) \\ a18 = 73 \\ a + 17d = 73 - - - - - (2) \\ on \:subtracting \: so(1) \: from \: (2) \\ 17d - 9d = 73 - 41 \\ 8d = 32 \\ d = \frac{32}{8} \\ d = 4 \\ putting \: value \: of \: d \: in \: (1) \\ a + 9 \times 4 = 41 \\ a = 41 - 36 \\ a = 5 \\ a26 = a + 25d \\ = 5 + 25 \times 4 \\ = 105$

$\huge{\boxed{\sf{ a26=105 }}}$

Answer 2

Given:-

• The 10th and 18th terms of an A.P. are 41 and 73 respectively.

To find:-

• Find 26th term...?

Solutions:-

• 10th term of Ap is 41
• 18th term of Ap is 73

We know that,

The 10th term of Ap is 41.

=> an = a + (n - 1)d

=> a10 = a + (10 - 1)d

=> 41 = a + 9d ..........(i).

The 18th term of Ap is 73

=> an = a + (n - 1)d

=> a18 = a + (18 - 1)d

=> 73 = a + 17d ..........(ii).

Now, Subtracting Eq. (ii) and (i) we get,

${a} \: + \: {17d} \: \: = \: \: {73} \\ {a} \: + \: {9d} \: \: = \: \: {41} \\ \underline{ - \: \: \: \: \: \: \: \: - \: \: \: \: \: \: \: \: = \: \: \: \: \: \: - \: \: \: \: \: \: \: \: \: } \\ \: \: \: \: \: \: \: \: {8d} \: \: \: \: \: \: \: \: = \: \: \: {32}$

=> d = 32/8

=> d = 4

Now, putting the value of y in Eq. (i).

=> a + 9d = 41

=> a + 9(4) = 41

=> a + 32 = 41

=> a = 41 - 32

=> a = 5

Thus, a = 5, d = 4, n = 26

Substitution the above value in the formula,

=> an = a + (n - 1)d

=> a26 = 5 + (26 - 1)(4)

=> a26 = 5 + (25)(4)

=> a26 = 5 + 100

=> a26 = 105

Hence, the 26th term of Ap is 105.