Question 4

A.P.: 3, 8, 13, 18, ... का कौन सा पद 78 है?

Given A.P.: 3, 8, 13, 18, ...

First term $a = 3$
Common difference $d = 8 - 3 = 5$

We need to find $n$ for which $a_n = 78$.

Using the nth term formula of an A.P.: $a_n = a + (n-1)d$ $78 = 3 + (n-1) \cdot 5$ $78 = 3 + 5n - 5$ $78 = 5n - 2$ $80 = 5n$ $n = \frac{80}{5} = 16$

Thus, the 16th term is 78.

Question 5

निम्नलिखित समांतर श्रेणियों में से प्रत्येक श्रेणी में कितने पद हैं?

(i) $7, 13, 19, \ldots, 205$

First term $a = 7$
Common difference $d = 13 - 7 = 6$
Last term $a_n = 205$

Using the nth term formula: $a_n = a + (n-1)d$ $205 = 7 + (n-1) \cdot 6$ $205 = 7 + 6n - 6$ $205 = 6n + 1$ $204 = 6n$ $n = \frac{204}{6} = 34$

So, there are 34 terms.

(ii) $18, 15\frac{1}{2}, 13, \ldots, -47$

First term $a = 18$
Common difference $d = 15\frac{1}{2} - 18 = -2.5$
Last term $a_n = -47$

Using the nth term formula: $a_n = a + (n-1)d$ $-47 = 18 + (n-1)(-2.5)$ $-47 = 18 - 2.5n + 2.5$ $-47 = 20.5 - 2.5n$ $-67.5 = -2.5n$ $n = \frac{67.5}{2.5} = 27$

So, there are 27 terms.

Question 6

क्या A.P. 11, 8, 5, 2, ... का एक पद -150 है? क्यों?

First term $a = 11$
Common difference $d = 8 - 11 = -3$

We need to find if there is any term $a_n$ such that $a_n = -150$.

Using the nth term formula: $a_n = a + (n-1)d$ $-150 = 11 + (n-1)(-3)$ $-150 = 11 - 3n + 3$ $-150 = 14 - 3n$ $-164 = -3n$ $n = \frac{164}{3} \approx 54.67$

Since $n$ is not an integer, $-150$ is not a term in the given A.P.

Question 7

उस A.P. का 31वाँ पद ज्ञात कीजिए, जिसका 11वाँ पद 38 है और 16वाँ पद 73 है।

Given: $a + 10d = 38$ $a + 15d = 73$

Subtracting the first equation from the second: $(a + 15d) - (a + 10d) = 73 - 38$ $5d = 35$ $d = 7$

Now, substituting $d$ back into the first equation: $a + 10 \cdot 7 = 38$ $a + 70 = 38$ $a = 38 - 70 = -32$

To find the 31st term: $a + 30d = -32 + 30 \cdot 7$ $= -32 + 210$ $= 178$

Thus, the 31st term is 178.

Question 8

एक A.P. में 50 पद हैं, जिसका तीसरा पद 12 है और अंतिम पद 106 है। इसका 29वाँ पद ज्ञात कीजिए।

Given: $a_3 = a + 2d = 12$ $a_{50} = a + 49d = 106$

From the first equation: $a + 2d = 12$ $a = 12 - 2d$

Substitute $a$ into the second equation: $(12 - 2d) + 49d = 106$ $12 + 47d = 106$ $47d = 94$ $d = 2$

Now, substituting $d$ back into $a$: $a = 12 - 2 \cdot 2$ $a = 12 - 4$ $a = 8$

To find the 29th term: $a_{29} = a + 28d$ $= 8 + 28 \cdot 2$ $= 8 + 56$ $= 64$

Thus, the 29th term is 64.

If you need further explanations or more detailed steps for these problems, please let me know. Here are some questions for you to consider:

1. How to derive the general formula for the nth term of an A.P.?
2. What is the process to find the number of terms in a given A.P.?
3. How to determine if a specific value is a term in an A.P.?
4. How to solve a problem involving multiple terms of an A.P.?
5. What are the practical applications of A.P. in real life?

Tip: When working with arithmetic progressions, always check the common difference and first term carefully, as these form the basis of all calculations.