Question

the sum of 5 th and 16 th term of an arithmetic sequence is 67. if the 10 th term is 32 what is it's 11 th term​

Answer 1

Answer:

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Step-by-step explanation:

so for the above ap we need to find t11 i.e. its 11th term

now let the 5th term of an AP be a+4d and 16th term be a+15d

so according to first condition

a+4d+a+15d=67

ie 2a+19d=67 (1)

so for second condition

we have to use formula

tn=a+(n-1)d

so according to second condition

t10=32

put n=10 in above formula

we get t10=a+(10-1)d

ie a+9d=32 (2)

so multiply (2) by 2

we get 2a+18d=64 (3)

subtract (3) from (2)

we get

2a+19d=67

2a+18d=64

(-) (-) (-)

__________

d=3

so d=3

substitute value of d in any of three equations

we get a=5

so now again using

tn=a+(n-1)d

but put n=11

we get t11=5+[(11-1)×3]

=5+(10×3)

=30+5

so t11=35

hence 11th term of above ap is 35