Question

In Ap Given a=5,d=3 the find a10 and a15​

Answer 1

${\huge{\mathfrak{\red{Solution}}}}$

${\huge{\bf{\orange{given}}}}$

$a=5\:and\:d=3$

${\bf{\pink{a10=>}}}$

$a10=a+9d$

$a10=5+9×3$

$a10=5+27=32$

${\bf{\blue{a15=>}}}$

$a15=a+14d$

$a15=5+14×3$

$a15=5+ 42$

$a15=47$

Answer 2

Given :

• First ferm ( a ) = 5

• Common difference ( d ) = 3

To Find :

• 10th term of A.P

• 15th term of A.P

Solution :

By using A.P formula

$\large\implies\boxed{\boxed{ \sf A_{10} =a + 9d}} \\ \\\implies \sf A_{10} =5 + 9 \times 3 \\ \\\implies \sf A_{10} =5 + 27 \\ \\\large\implies\boxed{\boxed{ \sf \green{ A_{10} =32}}}$

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$\large \implies\boxed{\boxed{ \sf A_{15} =a + 14d }}\\ \\ \implies \sf A_{15} =5 + 14 \times 3 \\ \\ \implies \sf A_{15} =5 + 42 \\ \\ \large \implies \boxed{\boxed{\sf \blue{A_{15}=47}}}$