Question

find Arthemtic seires if 5th term is 19 and S4 =a9 +1​

Answer 1

Answer:

Question:-

Find an AP series in which

$\bf \:a_5 = 19 \: and \: S_4 = a_9 + 1$

Answer

Given :-

$\bf \:a_5 = 19 \: and \: S_4 = a_9 + 1$

To Find :-

• An AP series.

Formula used :-

$\bf \:a_n = a + (n - 1) \times d$where,

• a = first term
• d = Common Difference
• n = number of terms

$\bf \: S_n = \dfrac{n}{2} (2a + (n - 1) \times d)$

where,

• a = first term
• d = Common Difference
• n = number of terms

Solution:-

Let the first term of an AP is 'a' and common difference be 'd'.

Step 1 :-

$\bf \:a_5 =19$

$\bf\implies \:a + 4d = 19$

$\bf\implies \:a = 19 - 4d........(1)$

Step 2 :-

$\bf\implies \:S_4 = a_9 + 1$

$\bf\implies \:\dfrac{4}{2} (2a + 3d) = a + 8d + 1$

$\bf\implies \:2(2a + 3d) = a + 8d + 1$

$\bf\implies \:4a + 6d = a + 8d + 1$

$\bf\implies \:3a - 2d = 1$

Put the value of a = 19 - 4d, we get

$\bf\implies \:3(19 - 4d) - 2d = 1$

$\bf\implies \:57 - 12d - 2d = 1$

$\bf\implies \:14d = 56$

$\bf\implies \:d = 4$

Put d = 4 in equation (1), we get

$\bf\implies \:a = 19 - 4 \times 4 = 19 - 16 = 3$

$\bf \:So, \: required \: AP \: series \: is \: 3 , \: 7, \: 11, \: 15,....$

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