Question

Sum of first 9 terms of an arithmetic sequence is 261.Sum first 14 terms is 661.
1)what is the 5th term of the sequence
2)what is the sum of first term and 14th term
3)what is the 10th term.
4)find the common difference
5)write the sequence

Answer 1

Answer:-

Given:

Sum of first nine terms of an AP = 261

Sum of first 14 terms = 661.

We know that,

$\sf{Sum \: of \: first \: n \: terms \: of \: an \: AP( S_{n}) = \frac{n}{2} [2a + (n - 1)d] }\\ \\ →\sf{ s_{9} = \frac{9}{2} [2a + (9 - 1)d]} \\ \\ → \sf{\frac{9}{2} [2(a + 4d)] = 261 }\\ \\→ \sf{a + 4d = \frac{261}{9} } \\ \\ →\sf{a + 4d = 29} \\ \\→\sf{ a = 29 - 4d \: - - equation \: (1)}$

Similarly,

$→\sf{\frac{14}{2} [2a + (14 - 1)d] = 661 }\\ \\ →\sf{ 7( 2a + 13d) = 661 } \\ \\→ \sf{14a + 91d = 661 }\\ \\ →\sf{14(29 - 4d) + 91d = 661} \\ \\ →\sf{406 - 56d + 91d = 661} \\ \\→\sf{ 35d = 661 - 406 }\\ \\→\sf{ 35d = 255 }\\ \\→\sf{ d = \frac{255}{35} } \\ \\ →\sf{d = \frac{51}{7} }$

Substitute "d" value in equation (1).

$→ \sf{a = 29 - 4d} \\ \\ →\sf{a = 29 - 4( \frac{51}{7} )} \\ \\ →\sf{a = \frac{203 - 204}{7} } \\ \\ →\sf{a = \frac{ - 1}{7} }$

1) We know that,

nth term of an AP = a + (n - 1)d

$→\sf{ a_{5} = a + 4d} \\ \\→ \sf{a _{5} = \frac{ - 1}{7} + 4( \frac{51}{7} )} \\ \\→ \sf{a_{5} = \frac{ - 1 + 204}{7}} \\ \\→\sf{ a_{5} = \frac{203}{7} }$

$2) \: \sf{a _{1} + a_{14} = a + a + 13d} \\ \\ →\sf{a_{1} + a_{14} = 2a + 13d }\\ \\→ \sf{ a_{1} + a _{14} = \frac{661}{7} }\\\\ 3) \: \sf{ a_{10} = a + 9d} \\ \\→ \sf{ a_{10} = \frac{ - 1}{7} + 9( \frac{51}{7} )} \\ \\ → \sf{a_{10} = \frac{ - 1 + 459}{7}} \\ \\ → \sf{a_{10} = \frac{458}{7} }\\\\4) \: \sf{d = \frac{51}{7} }$

$5) \: \sf{a = \frac{ - 1}{7}\: ; \: d = \frac{51}{7} } \\ \\→ \sf{ a_{2} = a + d} \\ \\→\sf{ a _{2} = \frac{ - 1 + 51}{7} = \frac{50}{7} } \\ \\→\sf{ a _{3} = a + d + d} \\ \\→ \sf{ a_{3} = \frac{50 + 51}{7} = \frac{101}{3} }$

Hence, The required AP is - 1/7 ; 50/7 ; 101/7...