Question

for
an. A.P., t7 = 10 and t13 = 34 find the value of t10​

Answer 1

Answer:

Step-by-step explanation:

t7 = 10

t + 6d = 10

t = 10-6d  ----(1)

t13 = 34

t + 12d = 34

from eq1

10 - 6d + 12d = 34

6d = 24

d = 4

t = 10 - 24 = -14

t10 = t + 9d

= -14 + 36

=22

Answer 2

Answer:

10th term of AP is t₁₀ = 22

Step-by-step explanation:

given that:

in an AP , t₇ = 10 = 7th term of an AP

t₁₃ = 34 = 13th term of an AP

we have to find the value of 10th term of AP = t₁₀

if we consider 1st term of AP as t₁,  tₙ can be written as t₁ + (n-1)d

where tₙ is nth term of AP and 'd' is difference of second and first term.

so , t₇ can be written as t₁ + (7 - 1)d

t₇= t₁ + (7 - 1)d = 10

t₁ + 6d = 10

t₁ = 10 - 6d

also , t₁₃= 34 = t₁ + (13 - 1)d

t₁ + 12d = 34

substitute value of 't₁ = 10 - 6d' in above equation:

10 - 6d + 12d = 34

10 + 6d = 34

6d = 34 - 10 = 24

d = 24/6

d = difference = 4

substitute value of d= 4 in the equation; t₁ + 6d = 10

t₁ + (6)(4) = 10

t₁ + 24 = 10

t₁ = 10 - 24 = -14

first term of AP is (-14)

t₁₀ = t₁ + 9d

= -14 + (9)(4)

= -14 + 36

= 22

10th term of AP is t₁₀ = 22