Question

(iv) given a = 15, S10= 125, find d and a10​

Answer 1

Step-by-step explanation:

a

3

=15,s

10

=125

a

3

=a+(3−1)d

15=a+2d

a=15−2d

s

10

=

2

10

[2a+(10−1)d]

125=5[2a+9d]

25=2a+9d

25=2(15−2d)+9d

25=30−4d+9d

5d=−5

d=−1

a=15−2×(−1)

a=17

a

10

=a+(10−1)(−1)

a

10

=17−9

a

10

=26

Answer 2

$\large \sf\underline{ \underline{ \red{given : }}}$

• a = 15

• S(10) = 135

• n = 10

$\\ \\ \large \sf\underline{ \underline{ \red{to \: find : }}}$

• Common difference (d)

• 10th term a(10)

$\\ \\ \large \sf\underline{ \underline{ \red{to \: know : }}}$

$\boxed{ \tt{s_{n} = \frac{n}{2}(1st \: term \: + last \: term }}$

$\boxed{ \tt{ a_{n} = a + (n - 1)d}}$

$\ \\ \large \sf\underline{ \underline{ \red{solution : }}}$

• Sum of first 10 terms is 125.

• 1st term is 15.

• There are 10 terms.

So, putting these values in formula ,

$\\ \sf{125 = \frac{ { \cancel{10}} \: ^{5} }{ { \cancel2} \: ^{1} } (15 + a_{10})} \\ \\ \\ \sf{125 = 5(15 + a_{10}) } \\ \\ \\ \sf{ \frac{ { \cancel{125}} \: ^{25} }{ { \cancel5} \: ^{1} } = 15 + a_{10} } \\ \\ \sf{25 = 15 + a_{10} } \\ \\ \\ \boxed{\sf{ \pink{a_{10} = 10 }}}$

And ,

• a(n) = a + (n-1)d

$\\ \sf{ a_{10} = 15 + (10 - 1)d} \\ \\ \sf{ 10 =15 + 9d } \\ \\ \sf{10 - 15 = 9d} \\ \\ \sf{9d = - 5} \\ \\ \boxed{{ \sf{{ \pink{d = \frac{ - 5}{9} }}}}}$

Hence , a10 is 10 and d is -5/9.