Question

find the 15th term of an ap whose 6th term is -10 and 10th term is -26

Answer 1

Answer:

${15}^{th} term = ( - 34)$

Step-by-step explanation:

${6}^{th} term = ( - 10) \\ a_{6} = ( - 10), \: n = 6 \\ a _{n} = a + (n - 1)d \\ a_{6} = a + (6 - 1)d \\ ( - 10) = a + 5d - eq1 \\ {10}^{th } term = ( - 26)\\ a_{10} = ( - 26), \: n = 10 \\ a _{n} = a + (n - 1)d \\ a_{10} = a + (10 - 1)d \\ ( - 26) = a + 9d - eq2 \\ Add \: eq1 \: and \: eq2 \\ \: \: \: \: a + 9d = ( - 26) \\ ± \: a ±5d = ( - 10) \\ - - - - - - - - - - \\ 4d = ( - 16) \\ d = \frac{( - 16)}{4} \\ d = ( - 4) \\ Substitute \: d=(-4) \: in \: eq1 \\ a + 5d = ( - 10) \\ a + 5( - 4) = ( - 10) \\ a + ( - 20) = ( - 10) \\ a - 20 = 10 \\ a = 10 + 20 \\ a = 30 \\ {15}^{th} term \\ a _{n} = a + (n - 1)d \\ a_{15} = 30 + (15 - 1)( - 4) \\ a _{15} = 30 + 14( - 4) \\ a _{15} = 30 + ( - 64) \\ a _{15} = 30 - 64 \\ a _{15} = ( - 34)$