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Answer 1 · 2018-01-11T17:53:09

#62, and 3772#.

Let, #a and d# be the #1^(st)# term and the common difference** of the

AP in question.

Let, #u_n and s_n# denote the #n^(th)# term and sum of the first #n# terms.

Then, #u_n=a+(n-1)d, s_n=n/2{2a+(n-1)d}#.

#u_6=20 rArr a+5d=20........................<<1>>#.

#u_11=35 rArr a+10d=35...................<<2>>#.

Solving #<<1>> and <<2>>, a=5, d=3#.

#:. u_20=a+19d=5+57=62.........................."Answer "1.#

Further, #sum_(n=10)^50u_n=u_10+u_11+...+u_50#,

#=u_1+u_2+...+u_9+u_10+u_11+...+u_50-{u_1+u_2+...+u_9},#

#=s_50-s_9#,

#=50/2{2(5)+(50-1)3}-9/2{2(5)+8(3)}#,

#=25{10+147}-9/2{10+24}#,

#=3925-153#,

#:. sum_(n=10)^50u_n=3772..............................................."Answer 2"#.