Let a be the first term and d be the common difference of the given AP. Then,
`T_(4) + T_(8) = 24 rArr (a+3d) + (a+7d) = 24`
`rArr 2a+ 10d = 24`
`rArr a +5d = 12 " " ….(i)`
`"And," T_(6) + T(10) = 44 rArr (a +5d) + (a +9d) = 44`
`rArr 2a +14d = 44`
`rArr a +7d = 22. " "...(ii)`
On solving (i) and (ii), we get a = -13 and d = 5.
`therefore` the sum of first 10 terms of the given AP is given by
`S_(10) = ((10)/(2)) * (2a+9d) " " ["using"S_(n) = (n)/(2) {(2a + (n-1)d}]`
` = 5 xx {2 xx (-13) + 9 xx5} = 5(-26 +45) = 5 xx 19 = 95.`
Hence, the sum of first 10 terms of the given AP is 95.
