Question

The last term of an AP with 10 terms whose second term is -23 and 3rd term is -35 is ________​

Answer 1

$\sf \purple{ \clubs \:Given : }$

• AP has 10 terms
• Second term of AP = - 23
• Third term of AP = - 35

$\sf \purple{ \clubs \:To \: find: }$

• Last term of AP which is 10th term .

$\sf \purple{ \clubs \:Solution: }$

Second term of AP = - 23

$\tt \: a_n = a + (n - 1)d \\ \tt \: a_2 = a + (2 - 1)d \\ \tt \: a_2 = a + d \: \\ \tt \: a + d = - 23 \longrightarrow \: (1)$

Third term of AP = - 35

$\tt \: a_n = a + (n - 1)d \\ \tt \: a_3 = a + (3 - 1)d \\ \tt \: a_3 = a + 2d \\ \tt \: a + 2d = - 35 \: \longrightarrow \: (2)$

From (1)

$\tt \: a + d \: = - 23 \\ \tt \: a = - 23 - d$

Putting value of a in (2)

$\tt - 23 - d + 2d = - 35 \\ \tt \: d = - 35 + 23 \\ \tt \: d = - 12$

Putting value of d in (1)

$\tt \: a + d = - 23 \\ \tt \: a + ( - 12) = - 23 \\ \tt \: a - 12 = - 23 \\ \tt a = - 23 + 12 \\ \tt a = - 11$

Now for finding last term

n = 10 , a = -11 and d = -12

$\tt \: a_n = a + (n - 1)d \\ \tt \: a_{10} = - 11 + (10 - 1) - 12 \\ \tt \: a_{10} = - 11 +( 9 \times - 12) \\ \tt \: a_{10} = - 11 +( - 108) \\ \tt \: a_{10} = - 11 - 108 \\ \tt \: a_{10} = - 119$

The last term of an AP with 10 terms whose second term is -23 and 3rd term is -35 is -119 .

Answer 2

Answer:

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