Question

find the 31st term of an AP whose 11th term is 38 and the 16 term is 73​

Answer 1

Answer:

$\large{\underline{\boxed{\sf a_{31} = 178}}}$

Step-by-step explanation:

Given that ;

The 11ᵗʰ term of the AP is 38.

$\tt \implies a + 10d = 38 \: \sf \: \: ....(i)$

And, the 16ᵗʰ term is 73.

$\tt \implies a + 15d = 73 \: \sf \: \: ....(ii)$

On Subtracting equation (i) from (ii),

$\tt \implies a + 15d - (a + 10d)= 73 - 38 \\\\\cancel a + 15d - \cancel a - 10d = 35 \\ \\ \tt \implies 5d = 35 \\ \\ \tt \implies d = \frac{35}{5} \\ \\ \tt \implies d = 7$

On substituting the value of d in (i);

$\tt \implies a + 10d = 38 \\ \\ \tt \implies a + 10 (7) = 38 \\ \\ \tt \implies a + 70 = 38 \\ \\ \tt \implies a = 38 - 70 \\ \\ \tt \implies a = - 32$

Now, we need to find the 31ˢᵗ term,

$\tt \implies a_n = a + (n - 1)d$

Here, we have

• n = 31
• a = - 32
• d = 7

$\tt \implies a_{31} = - 32+ (31 - 1) 7 \\ \\ \tt \implies a_{31} = -32 + 30 * 7 \\ \\ \tt \implies a_{31} = -32 + 210 \\ \\ \tt \implies a_{31} = 178$

Hence, the 31ˢᵗ term is 178.

Answer 2

Answer:

$\huge{\boxed{\sf{a31=178}}}$

Step-by-step explanation:

$\huge{\boxed{\sf{SOLUTION-}}}$

$\huge{\boxed{\sf{GIVEN-}}}$

$\huge{\boxed{\sf{a11=38}}}$

$a11 =a + 10d = 38 - - - - - (1)$

$\huge{\boxed{\sf{a16=73}}}$

$a16 = a + 15d = 73 - - - - - (2)$

$\huge{\boxed{\sf{a31=?}}}$

$on \:subtracting \:( 1) \: from \: (2) \: we \: get \\ = ) 5d = 35\\ = )d = \frac{35}{5} \\ =) d = 7\\ putting \: value \: of \: d \: in \: (1) \\ a + 10d = 38 \\ a + 10 \times 7 = 38 \\ a + 70 = 38 \\ a = 38 - 70 \\ a = - 32 \\ we \: have \: to \: find \: a31 \: so \\ a31 = a + 30d \\ \: \: \: \: \: \: \: = - 32 + 30 \times 7 \\ \: \: \: \: \: \: \: = -32+ 210 \\ \: \: \: \: \: \: \: = 178$

$\huge{\boxed{\sf{a31=178}}}$